Mixed mathematical chemical problem












3












$begingroup$



How to calculate the concentration of all species present in a solution with $mathrm{0.3~M}$ $mathrm{NaH_2PO_4}$?




The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species. The $mathrm{NaH_2PO_4}$ solution contains the ions $mathrm{Na^+}$, $mathrm{H^+}$, $mathrm{OH^−}$, $mathrm{H_2PO_4^{-}}$, $mathrm{HPO_4^{2-}}$, $mathrm{PO_4^{3-}}$ and undissociated acid $mathrm{H_3PO_4}$.



I form seven independent equations in order to specify the unknown concentrations of the seven species present in an aqueous solution of $mathrm{NaH_2PO_4}$. these equations are:



$$
K_1 = frac{[mathrm{H^+}][mathrm{H_2PO_4^-}]}{[mathrm{H_3PO_4}]} = 7.6 times 10^{−3} tag{1}$$

$$K_2 = frac{[mathrm{H^+}][mathrm{HPO_4^{2-}}]}{[mathrm{H_2PO_4^-}]} = 6.2 times 10^{−8} tag{2}$$
$$K_{3} = frac{[mathrm{H^+}][mathrm{PO_4^{3-}}]}{[mathrm{HPO_4^{2-}}]} = 2.1 times 10^{−13} tag{3}$$
$$K_mathrm{w} = {[mathrm{H^+}][mathrm{OH^-}]} = 1 times 10^{−14} label{eq:4}tag{4}
$$

$$C_mathrm{P} =[mathrm{Na^+}]= [mathrm{NaH_2PO_4}]_0 ={0.3}tag{5}$$
$$C_mathrm{P} = [mathrm{PO_4^{3-}}] + [mathrm{HPO_4^{2-}}] + [mathrm{H_2PO_4^-}] + [mathrm{H_3PO_4}] tag{6}$$
$${[mathrm{H_3PO_4}] + [mathrm{H^+}] = [mathrm{HPO_4^{2-}}] + 2[mathrm{PO_4^{3-}}] +frac{K_mathrm{w}}{[mathrm{H^+}]}}tag{7}$$



Neglecting $[mathrm{Na^+}]$ and equation $eqref{eq:4}$, then given $C_mathrm{P} = {0.3}$ and $K_mathrm{w}=1times10^{−14}$, how to solve five equations to find five variables using wolframalpha for solving quintic equation?










share|cite|improve this question











$endgroup$












  • $begingroup$
    @Claude Leibovici
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:11
















3












$begingroup$



How to calculate the concentration of all species present in a solution with $mathrm{0.3~M}$ $mathrm{NaH_2PO_4}$?




The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species. The $mathrm{NaH_2PO_4}$ solution contains the ions $mathrm{Na^+}$, $mathrm{H^+}$, $mathrm{OH^−}$, $mathrm{H_2PO_4^{-}}$, $mathrm{HPO_4^{2-}}$, $mathrm{PO_4^{3-}}$ and undissociated acid $mathrm{H_3PO_4}$.



I form seven independent equations in order to specify the unknown concentrations of the seven species present in an aqueous solution of $mathrm{NaH_2PO_4}$. these equations are:



$$
K_1 = frac{[mathrm{H^+}][mathrm{H_2PO_4^-}]}{[mathrm{H_3PO_4}]} = 7.6 times 10^{−3} tag{1}$$

$$K_2 = frac{[mathrm{H^+}][mathrm{HPO_4^{2-}}]}{[mathrm{H_2PO_4^-}]} = 6.2 times 10^{−8} tag{2}$$
$$K_{3} = frac{[mathrm{H^+}][mathrm{PO_4^{3-}}]}{[mathrm{HPO_4^{2-}}]} = 2.1 times 10^{−13} tag{3}$$
$$K_mathrm{w} = {[mathrm{H^+}][mathrm{OH^-}]} = 1 times 10^{−14} label{eq:4}tag{4}
$$

$$C_mathrm{P} =[mathrm{Na^+}]= [mathrm{NaH_2PO_4}]_0 ={0.3}tag{5}$$
$$C_mathrm{P} = [mathrm{PO_4^{3-}}] + [mathrm{HPO_4^{2-}}] + [mathrm{H_2PO_4^-}] + [mathrm{H_3PO_4}] tag{6}$$
$${[mathrm{H_3PO_4}] + [mathrm{H^+}] = [mathrm{HPO_4^{2-}}] + 2[mathrm{PO_4^{3-}}] +frac{K_mathrm{w}}{[mathrm{H^+}]}}tag{7}$$



Neglecting $[mathrm{Na^+}]$ and equation $eqref{eq:4}$, then given $C_mathrm{P} = {0.3}$ and $K_mathrm{w}=1times10^{−14}$, how to solve five equations to find five variables using wolframalpha for solving quintic equation?










share|cite|improve this question











$endgroup$












  • $begingroup$
    @Claude Leibovici
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:11














3












3








3





$begingroup$



How to calculate the concentration of all species present in a solution with $mathrm{0.3~M}$ $mathrm{NaH_2PO_4}$?




The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species. The $mathrm{NaH_2PO_4}$ solution contains the ions $mathrm{Na^+}$, $mathrm{H^+}$, $mathrm{OH^−}$, $mathrm{H_2PO_4^{-}}$, $mathrm{HPO_4^{2-}}$, $mathrm{PO_4^{3-}}$ and undissociated acid $mathrm{H_3PO_4}$.



I form seven independent equations in order to specify the unknown concentrations of the seven species present in an aqueous solution of $mathrm{NaH_2PO_4}$. these equations are:



$$
K_1 = frac{[mathrm{H^+}][mathrm{H_2PO_4^-}]}{[mathrm{H_3PO_4}]} = 7.6 times 10^{−3} tag{1}$$

$$K_2 = frac{[mathrm{H^+}][mathrm{HPO_4^{2-}}]}{[mathrm{H_2PO_4^-}]} = 6.2 times 10^{−8} tag{2}$$
$$K_{3} = frac{[mathrm{H^+}][mathrm{PO_4^{3-}}]}{[mathrm{HPO_4^{2-}}]} = 2.1 times 10^{−13} tag{3}$$
$$K_mathrm{w} = {[mathrm{H^+}][mathrm{OH^-}]} = 1 times 10^{−14} label{eq:4}tag{4}
$$

$$C_mathrm{P} =[mathrm{Na^+}]= [mathrm{NaH_2PO_4}]_0 ={0.3}tag{5}$$
$$C_mathrm{P} = [mathrm{PO_4^{3-}}] + [mathrm{HPO_4^{2-}}] + [mathrm{H_2PO_4^-}] + [mathrm{H_3PO_4}] tag{6}$$
$${[mathrm{H_3PO_4}] + [mathrm{H^+}] = [mathrm{HPO_4^{2-}}] + 2[mathrm{PO_4^{3-}}] +frac{K_mathrm{w}}{[mathrm{H^+}]}}tag{7}$$



Neglecting $[mathrm{Na^+}]$ and equation $eqref{eq:4}$, then given $C_mathrm{P} = {0.3}$ and $K_mathrm{w}=1times10^{−14}$, how to solve five equations to find five variables using wolframalpha for solving quintic equation?










share|cite|improve this question











$endgroup$





How to calculate the concentration of all species present in a solution with $mathrm{0.3~M}$ $mathrm{NaH_2PO_4}$?




The whole truth is that any time that you add any phosphate ion into an aqueous solution, then you will have all four phosphate species. The $mathrm{NaH_2PO_4}$ solution contains the ions $mathrm{Na^+}$, $mathrm{H^+}$, $mathrm{OH^−}$, $mathrm{H_2PO_4^{-}}$, $mathrm{HPO_4^{2-}}$, $mathrm{PO_4^{3-}}$ and undissociated acid $mathrm{H_3PO_4}$.



I form seven independent equations in order to specify the unknown concentrations of the seven species present in an aqueous solution of $mathrm{NaH_2PO_4}$. these equations are:



$$
K_1 = frac{[mathrm{H^+}][mathrm{H_2PO_4^-}]}{[mathrm{H_3PO_4}]} = 7.6 times 10^{−3} tag{1}$$

$$K_2 = frac{[mathrm{H^+}][mathrm{HPO_4^{2-}}]}{[mathrm{H_2PO_4^-}]} = 6.2 times 10^{−8} tag{2}$$
$$K_{3} = frac{[mathrm{H^+}][mathrm{PO_4^{3-}}]}{[mathrm{HPO_4^{2-}}]} = 2.1 times 10^{−13} tag{3}$$
$$K_mathrm{w} = {[mathrm{H^+}][mathrm{OH^-}]} = 1 times 10^{−14} label{eq:4}tag{4}
$$

$$C_mathrm{P} =[mathrm{Na^+}]= [mathrm{NaH_2PO_4}]_0 ={0.3}tag{5}$$
$$C_mathrm{P} = [mathrm{PO_4^{3-}}] + [mathrm{HPO_4^{2-}}] + [mathrm{H_2PO_4^-}] + [mathrm{H_3PO_4}] tag{6}$$
$${[mathrm{H_3PO_4}] + [mathrm{H^+}] = [mathrm{HPO_4^{2-}}] + 2[mathrm{PO_4^{3-}}] +frac{K_mathrm{w}}{[mathrm{H^+}]}}tag{7}$$



Neglecting $[mathrm{Na^+}]$ and equation $eqref{eq:4}$, then given $C_mathrm{P} = {0.3}$ and $K_mathrm{w}=1times10^{−14}$, how to solve five equations to find five variables using wolframalpha for solving quintic equation?







systems-of-equations nonlinear-system chemistry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 18 '18 at 15:54









andselisk

264311




264311










asked Dec 5 '18 at 12:10









Adnan AL-AmlehAdnan AL-Amleh

1163




1163












  • $begingroup$
    @Claude Leibovici
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:11


















  • $begingroup$
    @Claude Leibovici
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:11
















$begingroup$
@Claude Leibovici
$endgroup$
– Adnan AL-Amleh
Dec 5 '18 at 12:11




$begingroup$
@Claude Leibovici
$endgroup$
– Adnan AL-Amleh
Dec 5 '18 at 12:11










2 Answers
2






active

oldest

votes


















2












$begingroup$

Too long for a comment. This just deals with the numerical aspects of the problem.



Starting from @Batominovski answer, let us define both sides of the charge balance.
$$text{lhs}=y^5+(C_P+K_1),y^4+K_1K_2,y^3+K_1K_2K_3,y^2$$
$$text{rhs}=K_w,y^3+(K_1K_w+C_PK_1K_2),y^2+(K_1K_2K_w+2C_PK_1K_2K_3),y+K_1K_2K_3K_w$$
Instead of solving $text{lhs}=text{rhs}$ for $y$, let $y=10^{-text{pH}}$ and consider that we look for the zero of function
$$f(text{pH})=log(text{lhs})-log(text{rhs})$$ which, if plooted, looks quite linear.



As a proof, using a Taylor expansion built at $text{pH}=7$ (corresponding to $ y=sqrt{k_w}$), the first estimate is $text{pH}=4.65444$ while the exact solution is $text{pH}=4.66886$. One more iteration of Newton method and we are done (as shown below for a ridiculous number of significant figures)
$$left(
begin{array}{cc}
n & text{pH}_n \
0 & 7.00000000000 \
1 & 4.65444138638 \
2 & 4.66886040691 \
3 & 4.66886042380
end{array}
right)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Claudi leibovici : I got the same as your answer by WolframAlpha with the appreciation of the powerful effort and noble interest.
    $endgroup$
    – Adnan AL-Amleh
    Dec 7 '18 at 16:10



















2












$begingroup$

For simplicity, write $x_0:=[texttt{H}_3texttt{PO}_4]$, $x_1:=[texttt{H}_2texttt{PO}_4^-]$, $x_2:=[texttt{H}texttt{PO}_4^{2-}]$, and $x_3:=[texttt{PO}_4^{3-}]$ so that $x_i$ is the concentration of the phosphate species with negative-$i$ charge, for $i=0,1,2,3$. Let $y$ denote $[texttt{H}^+]$. Therefore,
$$x_0+x_1+x_2+x_3=C_P,,$$
$$x_0+y=x_2+2x_3+frac{K_w}{y},,$$
$$K_1=frac{yx_1}{x_0},,$$
$$K_2=frac{yx_2}{x_1},,$$
and
$$K_3=frac{yx_3}{x_2},.$$
This gives $x_1=dfrac{K_1x_0}{y}$, $x_2=dfrac{K_2x_1}{y}=dfrac{K_1K_2x_0}{y^2}$, and $x_3=dfrac{K_3x_2}{y}=dfrac{K_1K_2K_3x_0}{y^3}$. Hence, we end up with two equations in $x_0$ and $y$:
$$x_0+frac{K_1x_0}{y}+frac{K_1K_2x_0}{y^2}+frac{K_1K_2K_3x_0}{y^3}=C_Ptag{*}$$
and
$$x_0+y=frac{K_1K_2x_0}{y^2}+frac{2K_1K_2K_3x_0}{y^3}+frac{K_w}{y},.tag{#}$$
From (*), we easily get
$$x_0=frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},.$$
Plugging this into (#) yields
$$frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},left(1-frac{K_1K_2}{y^2}-frac{2K_1K_2K_3}{y^3}right)+y-frac{K_w}{y}=0,.$$
That is,
$$y^5+(C_P+K_1)y^4+(K_1K_2-K_w)y^3-(K_1K_w+C_PK_1K_2-K_1K_2K_3)y^2-(K_1K_2K_w+2C_PK_1K_2K_3)y-K_1K_2K_3K_w=0,.$$
I think you need a numerical solver to solve for $y$ from the last equation. Once you know $y$, you can find the values of $x_0$, $x_1$, $x_2$, and $x_3$.



Mathematica tells me that there is only one positive real solution $y$ (or using Descartes's Rule of Signs, you know that there exists only one positive real solution) to the previous equation, which is $yapprox 2.1times 10^{-5}$. This means $x_0approx 8.4times 10^{-4}$, $x_1approx 0.30$, $x_2approx 8.6times 10^{-4}$, and $x_3approx 8.5times 10^{-12}$. In a sense, you should expect $x_1approx C_P=0.30$. After all, the dissociative reactions into other phosphate species are quite weak.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks, I appreciate to derive a quintic equation to compare it with my equation.
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:53











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026989%2fmixed-mathematical-chemical-problem%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

Too long for a comment. This just deals with the numerical aspects of the problem.



Starting from @Batominovski answer, let us define both sides of the charge balance.
$$text{lhs}=y^5+(C_P+K_1),y^4+K_1K_2,y^3+K_1K_2K_3,y^2$$
$$text{rhs}=K_w,y^3+(K_1K_w+C_PK_1K_2),y^2+(K_1K_2K_w+2C_PK_1K_2K_3),y+K_1K_2K_3K_w$$
Instead of solving $text{lhs}=text{rhs}$ for $y$, let $y=10^{-text{pH}}$ and consider that we look for the zero of function
$$f(text{pH})=log(text{lhs})-log(text{rhs})$$ which, if plooted, looks quite linear.



As a proof, using a Taylor expansion built at $text{pH}=7$ (corresponding to $ y=sqrt{k_w}$), the first estimate is $text{pH}=4.65444$ while the exact solution is $text{pH}=4.66886$. One more iteration of Newton method and we are done (as shown below for a ridiculous number of significant figures)
$$left(
begin{array}{cc}
n & text{pH}_n \
0 & 7.00000000000 \
1 & 4.65444138638 \
2 & 4.66886040691 \
3 & 4.66886042380
end{array}
right)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Claudi leibovici : I got the same as your answer by WolframAlpha with the appreciation of the powerful effort and noble interest.
    $endgroup$
    – Adnan AL-Amleh
    Dec 7 '18 at 16:10
















2












$begingroup$

Too long for a comment. This just deals with the numerical aspects of the problem.



Starting from @Batominovski answer, let us define both sides of the charge balance.
$$text{lhs}=y^5+(C_P+K_1),y^4+K_1K_2,y^3+K_1K_2K_3,y^2$$
$$text{rhs}=K_w,y^3+(K_1K_w+C_PK_1K_2),y^2+(K_1K_2K_w+2C_PK_1K_2K_3),y+K_1K_2K_3K_w$$
Instead of solving $text{lhs}=text{rhs}$ for $y$, let $y=10^{-text{pH}}$ and consider that we look for the zero of function
$$f(text{pH})=log(text{lhs})-log(text{rhs})$$ which, if plooted, looks quite linear.



As a proof, using a Taylor expansion built at $text{pH}=7$ (corresponding to $ y=sqrt{k_w}$), the first estimate is $text{pH}=4.65444$ while the exact solution is $text{pH}=4.66886$. One more iteration of Newton method and we are done (as shown below for a ridiculous number of significant figures)
$$left(
begin{array}{cc}
n & text{pH}_n \
0 & 7.00000000000 \
1 & 4.65444138638 \
2 & 4.66886040691 \
3 & 4.66886042380
end{array}
right)$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    @Claudi leibovici : I got the same as your answer by WolframAlpha with the appreciation of the powerful effort and noble interest.
    $endgroup$
    – Adnan AL-Amleh
    Dec 7 '18 at 16:10














2












2








2





$begingroup$

Too long for a comment. This just deals with the numerical aspects of the problem.



Starting from @Batominovski answer, let us define both sides of the charge balance.
$$text{lhs}=y^5+(C_P+K_1),y^4+K_1K_2,y^3+K_1K_2K_3,y^2$$
$$text{rhs}=K_w,y^3+(K_1K_w+C_PK_1K_2),y^2+(K_1K_2K_w+2C_PK_1K_2K_3),y+K_1K_2K_3K_w$$
Instead of solving $text{lhs}=text{rhs}$ for $y$, let $y=10^{-text{pH}}$ and consider that we look for the zero of function
$$f(text{pH})=log(text{lhs})-log(text{rhs})$$ which, if plooted, looks quite linear.



As a proof, using a Taylor expansion built at $text{pH}=7$ (corresponding to $ y=sqrt{k_w}$), the first estimate is $text{pH}=4.65444$ while the exact solution is $text{pH}=4.66886$. One more iteration of Newton method and we are done (as shown below for a ridiculous number of significant figures)
$$left(
begin{array}{cc}
n & text{pH}_n \
0 & 7.00000000000 \
1 & 4.65444138638 \
2 & 4.66886040691 \
3 & 4.66886042380
end{array}
right)$$






share|cite|improve this answer









$endgroup$



Too long for a comment. This just deals with the numerical aspects of the problem.



Starting from @Batominovski answer, let us define both sides of the charge balance.
$$text{lhs}=y^5+(C_P+K_1),y^4+K_1K_2,y^3+K_1K_2K_3,y^2$$
$$text{rhs}=K_w,y^3+(K_1K_w+C_PK_1K_2),y^2+(K_1K_2K_w+2C_PK_1K_2K_3),y+K_1K_2K_3K_w$$
Instead of solving $text{lhs}=text{rhs}$ for $y$, let $y=10^{-text{pH}}$ and consider that we look for the zero of function
$$f(text{pH})=log(text{lhs})-log(text{rhs})$$ which, if plooted, looks quite linear.



As a proof, using a Taylor expansion built at $text{pH}=7$ (corresponding to $ y=sqrt{k_w}$), the first estimate is $text{pH}=4.65444$ while the exact solution is $text{pH}=4.66886$. One more iteration of Newton method and we are done (as shown below for a ridiculous number of significant figures)
$$left(
begin{array}{cc}
n & text{pH}_n \
0 & 7.00000000000 \
1 & 4.65444138638 \
2 & 4.66886040691 \
3 & 4.66886042380
end{array}
right)$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 7 '18 at 11:27









Claude LeiboviciClaude Leibovici

123k1157134




123k1157134












  • $begingroup$
    @Claudi leibovici : I got the same as your answer by WolframAlpha with the appreciation of the powerful effort and noble interest.
    $endgroup$
    – Adnan AL-Amleh
    Dec 7 '18 at 16:10


















  • $begingroup$
    @Claudi leibovici : I got the same as your answer by WolframAlpha with the appreciation of the powerful effort and noble interest.
    $endgroup$
    – Adnan AL-Amleh
    Dec 7 '18 at 16:10
















$begingroup$
@Claudi leibovici : I got the same as your answer by WolframAlpha with the appreciation of the powerful effort and noble interest.
$endgroup$
– Adnan AL-Amleh
Dec 7 '18 at 16:10




$begingroup$
@Claudi leibovici : I got the same as your answer by WolframAlpha with the appreciation of the powerful effort and noble interest.
$endgroup$
– Adnan AL-Amleh
Dec 7 '18 at 16:10











2












$begingroup$

For simplicity, write $x_0:=[texttt{H}_3texttt{PO}_4]$, $x_1:=[texttt{H}_2texttt{PO}_4^-]$, $x_2:=[texttt{H}texttt{PO}_4^{2-}]$, and $x_3:=[texttt{PO}_4^{3-}]$ so that $x_i$ is the concentration of the phosphate species with negative-$i$ charge, for $i=0,1,2,3$. Let $y$ denote $[texttt{H}^+]$. Therefore,
$$x_0+x_1+x_2+x_3=C_P,,$$
$$x_0+y=x_2+2x_3+frac{K_w}{y},,$$
$$K_1=frac{yx_1}{x_0},,$$
$$K_2=frac{yx_2}{x_1},,$$
and
$$K_3=frac{yx_3}{x_2},.$$
This gives $x_1=dfrac{K_1x_0}{y}$, $x_2=dfrac{K_2x_1}{y}=dfrac{K_1K_2x_0}{y^2}$, and $x_3=dfrac{K_3x_2}{y}=dfrac{K_1K_2K_3x_0}{y^3}$. Hence, we end up with two equations in $x_0$ and $y$:
$$x_0+frac{K_1x_0}{y}+frac{K_1K_2x_0}{y^2}+frac{K_1K_2K_3x_0}{y^3}=C_Ptag{*}$$
and
$$x_0+y=frac{K_1K_2x_0}{y^2}+frac{2K_1K_2K_3x_0}{y^3}+frac{K_w}{y},.tag{#}$$
From (*), we easily get
$$x_0=frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},.$$
Plugging this into (#) yields
$$frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},left(1-frac{K_1K_2}{y^2}-frac{2K_1K_2K_3}{y^3}right)+y-frac{K_w}{y}=0,.$$
That is,
$$y^5+(C_P+K_1)y^4+(K_1K_2-K_w)y^3-(K_1K_w+C_PK_1K_2-K_1K_2K_3)y^2-(K_1K_2K_w+2C_PK_1K_2K_3)y-K_1K_2K_3K_w=0,.$$
I think you need a numerical solver to solve for $y$ from the last equation. Once you know $y$, you can find the values of $x_0$, $x_1$, $x_2$, and $x_3$.



Mathematica tells me that there is only one positive real solution $y$ (or using Descartes's Rule of Signs, you know that there exists only one positive real solution) to the previous equation, which is $yapprox 2.1times 10^{-5}$. This means $x_0approx 8.4times 10^{-4}$, $x_1approx 0.30$, $x_2approx 8.6times 10^{-4}$, and $x_3approx 8.5times 10^{-12}$. In a sense, you should expect $x_1approx C_P=0.30$. After all, the dissociative reactions into other phosphate species are quite weak.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks, I appreciate to derive a quintic equation to compare it with my equation.
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:53
















2












$begingroup$

For simplicity, write $x_0:=[texttt{H}_3texttt{PO}_4]$, $x_1:=[texttt{H}_2texttt{PO}_4^-]$, $x_2:=[texttt{H}texttt{PO}_4^{2-}]$, and $x_3:=[texttt{PO}_4^{3-}]$ so that $x_i$ is the concentration of the phosphate species with negative-$i$ charge, for $i=0,1,2,3$. Let $y$ denote $[texttt{H}^+]$. Therefore,
$$x_0+x_1+x_2+x_3=C_P,,$$
$$x_0+y=x_2+2x_3+frac{K_w}{y},,$$
$$K_1=frac{yx_1}{x_0},,$$
$$K_2=frac{yx_2}{x_1},,$$
and
$$K_3=frac{yx_3}{x_2},.$$
This gives $x_1=dfrac{K_1x_0}{y}$, $x_2=dfrac{K_2x_1}{y}=dfrac{K_1K_2x_0}{y^2}$, and $x_3=dfrac{K_3x_2}{y}=dfrac{K_1K_2K_3x_0}{y^3}$. Hence, we end up with two equations in $x_0$ and $y$:
$$x_0+frac{K_1x_0}{y}+frac{K_1K_2x_0}{y^2}+frac{K_1K_2K_3x_0}{y^3}=C_Ptag{*}$$
and
$$x_0+y=frac{K_1K_2x_0}{y^2}+frac{2K_1K_2K_3x_0}{y^3}+frac{K_w}{y},.tag{#}$$
From (*), we easily get
$$x_0=frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},.$$
Plugging this into (#) yields
$$frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},left(1-frac{K_1K_2}{y^2}-frac{2K_1K_2K_3}{y^3}right)+y-frac{K_w}{y}=0,.$$
That is,
$$y^5+(C_P+K_1)y^4+(K_1K_2-K_w)y^3-(K_1K_w+C_PK_1K_2-K_1K_2K_3)y^2-(K_1K_2K_w+2C_PK_1K_2K_3)y-K_1K_2K_3K_w=0,.$$
I think you need a numerical solver to solve for $y$ from the last equation. Once you know $y$, you can find the values of $x_0$, $x_1$, $x_2$, and $x_3$.



Mathematica tells me that there is only one positive real solution $y$ (or using Descartes's Rule of Signs, you know that there exists only one positive real solution) to the previous equation, which is $yapprox 2.1times 10^{-5}$. This means $x_0approx 8.4times 10^{-4}$, $x_1approx 0.30$, $x_2approx 8.6times 10^{-4}$, and $x_3approx 8.5times 10^{-12}$. In a sense, you should expect $x_1approx C_P=0.30$. After all, the dissociative reactions into other phosphate species are quite weak.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks, I appreciate to derive a quintic equation to compare it with my equation.
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:53














2












2








2





$begingroup$

For simplicity, write $x_0:=[texttt{H}_3texttt{PO}_4]$, $x_1:=[texttt{H}_2texttt{PO}_4^-]$, $x_2:=[texttt{H}texttt{PO}_4^{2-}]$, and $x_3:=[texttt{PO}_4^{3-}]$ so that $x_i$ is the concentration of the phosphate species with negative-$i$ charge, for $i=0,1,2,3$. Let $y$ denote $[texttt{H}^+]$. Therefore,
$$x_0+x_1+x_2+x_3=C_P,,$$
$$x_0+y=x_2+2x_3+frac{K_w}{y},,$$
$$K_1=frac{yx_1}{x_0},,$$
$$K_2=frac{yx_2}{x_1},,$$
and
$$K_3=frac{yx_3}{x_2},.$$
This gives $x_1=dfrac{K_1x_0}{y}$, $x_2=dfrac{K_2x_1}{y}=dfrac{K_1K_2x_0}{y^2}$, and $x_3=dfrac{K_3x_2}{y}=dfrac{K_1K_2K_3x_0}{y^3}$. Hence, we end up with two equations in $x_0$ and $y$:
$$x_0+frac{K_1x_0}{y}+frac{K_1K_2x_0}{y^2}+frac{K_1K_2K_3x_0}{y^3}=C_Ptag{*}$$
and
$$x_0+y=frac{K_1K_2x_0}{y^2}+frac{2K_1K_2K_3x_0}{y^3}+frac{K_w}{y},.tag{#}$$
From (*), we easily get
$$x_0=frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},.$$
Plugging this into (#) yields
$$frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},left(1-frac{K_1K_2}{y^2}-frac{2K_1K_2K_3}{y^3}right)+y-frac{K_w}{y}=0,.$$
That is,
$$y^5+(C_P+K_1)y^4+(K_1K_2-K_w)y^3-(K_1K_w+C_PK_1K_2-K_1K_2K_3)y^2-(K_1K_2K_w+2C_PK_1K_2K_3)y-K_1K_2K_3K_w=0,.$$
I think you need a numerical solver to solve for $y$ from the last equation. Once you know $y$, you can find the values of $x_0$, $x_1$, $x_2$, and $x_3$.



Mathematica tells me that there is only one positive real solution $y$ (or using Descartes's Rule of Signs, you know that there exists only one positive real solution) to the previous equation, which is $yapprox 2.1times 10^{-5}$. This means $x_0approx 8.4times 10^{-4}$, $x_1approx 0.30$, $x_2approx 8.6times 10^{-4}$, and $x_3approx 8.5times 10^{-12}$. In a sense, you should expect $x_1approx C_P=0.30$. After all, the dissociative reactions into other phosphate species are quite weak.






share|cite|improve this answer











$endgroup$



For simplicity, write $x_0:=[texttt{H}_3texttt{PO}_4]$, $x_1:=[texttt{H}_2texttt{PO}_4^-]$, $x_2:=[texttt{H}texttt{PO}_4^{2-}]$, and $x_3:=[texttt{PO}_4^{3-}]$ so that $x_i$ is the concentration of the phosphate species with negative-$i$ charge, for $i=0,1,2,3$. Let $y$ denote $[texttt{H}^+]$. Therefore,
$$x_0+x_1+x_2+x_3=C_P,,$$
$$x_0+y=x_2+2x_3+frac{K_w}{y},,$$
$$K_1=frac{yx_1}{x_0},,$$
$$K_2=frac{yx_2}{x_1},,$$
and
$$K_3=frac{yx_3}{x_2},.$$
This gives $x_1=dfrac{K_1x_0}{y}$, $x_2=dfrac{K_2x_1}{y}=dfrac{K_1K_2x_0}{y^2}$, and $x_3=dfrac{K_3x_2}{y}=dfrac{K_1K_2K_3x_0}{y^3}$. Hence, we end up with two equations in $x_0$ and $y$:
$$x_0+frac{K_1x_0}{y}+frac{K_1K_2x_0}{y^2}+frac{K_1K_2K_3x_0}{y^3}=C_Ptag{*}$$
and
$$x_0+y=frac{K_1K_2x_0}{y^2}+frac{2K_1K_2K_3x_0}{y^3}+frac{K_w}{y},.tag{#}$$
From (*), we easily get
$$x_0=frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},.$$
Plugging this into (#) yields
$$frac{C_P}{1+frac{K_1}{y}+frac{K_1K_2}{y^2}+frac{K_1K_2K_3}{y^3}},left(1-frac{K_1K_2}{y^2}-frac{2K_1K_2K_3}{y^3}right)+y-frac{K_w}{y}=0,.$$
That is,
$$y^5+(C_P+K_1)y^4+(K_1K_2-K_w)y^3-(K_1K_w+C_PK_1K_2-K_1K_2K_3)y^2-(K_1K_2K_w+2C_PK_1K_2K_3)y-K_1K_2K_3K_w=0,.$$
I think you need a numerical solver to solve for $y$ from the last equation. Once you know $y$, you can find the values of $x_0$, $x_1$, $x_2$, and $x_3$.



Mathematica tells me that there is only one positive real solution $y$ (or using Descartes's Rule of Signs, you know that there exists only one positive real solution) to the previous equation, which is $yapprox 2.1times 10^{-5}$. This means $x_0approx 8.4times 10^{-4}$, $x_1approx 0.30$, $x_2approx 8.6times 10^{-4}$, and $x_3approx 8.5times 10^{-12}$. In a sense, you should expect $x_1approx C_P=0.30$. After all, the dissociative reactions into other phosphate species are quite weak.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 19 '18 at 0:04

























answered Dec 5 '18 at 12:41









BatominovskiBatominovski

33.1k33293




33.1k33293












  • $begingroup$
    thanks, I appreciate to derive a quintic equation to compare it with my equation.
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:53


















  • $begingroup$
    thanks, I appreciate to derive a quintic equation to compare it with my equation.
    $endgroup$
    – Adnan AL-Amleh
    Dec 5 '18 at 12:53
















$begingroup$
thanks, I appreciate to derive a quintic equation to compare it with my equation.
$endgroup$
– Adnan AL-Amleh
Dec 5 '18 at 12:53




$begingroup$
thanks, I appreciate to derive a quintic equation to compare it with my equation.
$endgroup$
– Adnan AL-Amleh
Dec 5 '18 at 12:53


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3026989%2fmixed-mathematical-chemical-problem%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

How to change which sound is reproduced for terminal bell?

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Can I use Tabulator js library in my java Spring + Thymeleaf project?