How to derive one equation including only one variable from five equations?












0














The variables: H, M, X,Y,Z



The five equations:
$$
7.6times{10^{-3}} =frac{X H}{M} tag{1}$$

$$6.2times{10^{-8}} =frac{Y H}{X} tag{2}$$
$$2.1times{10^{-13}} =frac{Z H}{Y} tag{3}$$
$$H^2 +0.3H = X H + 2Y H + 3ZH+10^{-14}tag{4} $$
$$0.3 = M + X + Y + Ztag{5} $$
Quistion : Derive equation containing just H variable.










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  • Eliminate $M,X,Y,Z$ successively (naively) gives a degree 16 equation. This doesn't look like precalculus.
    – user10354138
    Dec 1 '18 at 14:07












  • @user10354138Can the degree 16 equation be solved by computer? How?
    – Adnan AL-Amleh
    Dec 1 '18 at 15:30
















0














The variables: H, M, X,Y,Z



The five equations:
$$
7.6times{10^{-3}} =frac{X H}{M} tag{1}$$

$$6.2times{10^{-8}} =frac{Y H}{X} tag{2}$$
$$2.1times{10^{-13}} =frac{Z H}{Y} tag{3}$$
$$H^2 +0.3H = X H + 2Y H + 3ZH+10^{-14}tag{4} $$
$$0.3 = M + X + Y + Ztag{5} $$
Quistion : Derive equation containing just H variable.










share|cite|improve this question






















  • Eliminate $M,X,Y,Z$ successively (naively) gives a degree 16 equation. This doesn't look like precalculus.
    – user10354138
    Dec 1 '18 at 14:07












  • @user10354138Can the degree 16 equation be solved by computer? How?
    – Adnan AL-Amleh
    Dec 1 '18 at 15:30














0












0








0







The variables: H, M, X,Y,Z



The five equations:
$$
7.6times{10^{-3}} =frac{X H}{M} tag{1}$$

$$6.2times{10^{-8}} =frac{Y H}{X} tag{2}$$
$$2.1times{10^{-13}} =frac{Z H}{Y} tag{3}$$
$$H^2 +0.3H = X H + 2Y H + 3ZH+10^{-14}tag{4} $$
$$0.3 = M + X + Y + Ztag{5} $$
Quistion : Derive equation containing just H variable.










share|cite|improve this question













The variables: H, M, X,Y,Z



The five equations:
$$
7.6times{10^{-3}} =frac{X H}{M} tag{1}$$

$$6.2times{10^{-8}} =frac{Y H}{X} tag{2}$$
$$2.1times{10^{-13}} =frac{Z H}{Y} tag{3}$$
$$H^2 +0.3H = X H + 2Y H + 3ZH+10^{-14}tag{4} $$
$$0.3 = M + X + Y + Ztag{5} $$
Quistion : Derive equation containing just H variable.







algebra-precalculus






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asked Dec 1 '18 at 13:57









Adnan AL-Amleh

1163




1163












  • Eliminate $M,X,Y,Z$ successively (naively) gives a degree 16 equation. This doesn't look like precalculus.
    – user10354138
    Dec 1 '18 at 14:07












  • @user10354138Can the degree 16 equation be solved by computer? How?
    – Adnan AL-Amleh
    Dec 1 '18 at 15:30


















  • Eliminate $M,X,Y,Z$ successively (naively) gives a degree 16 equation. This doesn't look like precalculus.
    – user10354138
    Dec 1 '18 at 14:07












  • @user10354138Can the degree 16 equation be solved by computer? How?
    – Adnan AL-Amleh
    Dec 1 '18 at 15:30
















Eliminate $M,X,Y,Z$ successively (naively) gives a degree 16 equation. This doesn't look like precalculus.
– user10354138
Dec 1 '18 at 14:07






Eliminate $M,X,Y,Z$ successively (naively) gives a degree 16 equation. This doesn't look like precalculus.
– user10354138
Dec 1 '18 at 14:07














@user10354138Can the degree 16 equation be solved by computer? How?
– Adnan AL-Amleh
Dec 1 '18 at 15:30




@user10354138Can the degree 16 equation be solved by computer? How?
– Adnan AL-Amleh
Dec 1 '18 at 15:30










1 Answer
1






active

oldest

votes


















1














This seems to be a chemical problem. When I saw it, I suspected possible traps because of the very small numbers and I rewrote the equations as
$$a =frac{X H}{M} tag{1}$$
$$b =frac{Y H}{X} tag{2}$$
$$c =frac{Z H}{Y} tag{3}$$
$$H^2 +frac 3 {10}H = X H + 2Y H + 3ZH+dtag{4} $$
$$frac 3 {10}= M + X + Y + Ztag{5} $$



In a first step, as @user10354138 commented, I used equations $(1,2,3,5)$ to express $X,Y,Z,M$ as functions of $H$. This leads to
$$X=frac{3 a H^2}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad Y=frac{3 a b H}{10 left(a b c+a b H+a H^2+H^3right)}$$
$$Z=frac{3 a b c}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad M=frac{3 H^3}{10 left(a b c+a b H+a H^2+H^3right)}$$
Now, replace these expressions in $(4)$, reduce to same denominator (which we shall ignore assuming it is non-zero) and put everything on the same side. We shall get
$$-10 a b c d-2 a b (3 c+5 d)H+a (b (10 c-3)-10 d)H^2+10 (a b-d)H^3+(10 a+3)
H^4+10 H^5=0tag 6$$
This is just a polynomial of degree $5$ which cannot be solved analytically and for which numerical methods will be required.



Replace $a,b,c,d$ by their values; you will get very ugly values for the coefficients.



If you want to solve $(6)$ for $H$, since we know the usual ranges of $H$, use inspection with $H=10^{-pH}$. Plotting the function as a function of $pH$, you will notice a very fast decreasing function with a solution close to $pH=4$. So, start Newton iterations using $pH_0=4$. You should get the following iterates
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.10601 \
2 & 4.21036 \
3 & 4.31193 \
4 & 4.40889 \
5 & 4.49815 \
6 & 4.57469 \
7 & 4.63121 \
8 & 4.66103 \
9 & 4.66846 \
10 & 4.66886
end{array}
right)$$
that is to say $H=10^{-4.66886}=0.0000214358$.



Edit



Plotting the function given by $(6)$ is not very good. You can have a much better perception using $H=10^{-pH}$ and plot, as a function of $pH$
$$f(pH)=frac{H^2 +frac 3 {10}H } {X H + 2Y H + 3ZH+d }-1tag 7$$ and look for its zero. Zoom once around the solution to get an estimate of $4.7$.



Solving $(7)$, using as before $pH_0=4$, Newton iterations would be
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.39558 \
2 & 4.63729 \
3 & 4.66879 \
4 & 4.66886
end{array}
right)$$
which is much faster than the previous one.






share|cite|improve this answer























  • I appreciate your interest and your effort in helping for solving the math part of the chemical problem posted in Chem.Stack Exchange how do I calculate the concentration of all species present in a solution with$pu{ 0.3M}ce{ NaH2PO4}$: So the concentration of$ce{ H3PO4, H2PO4 -, HPO4 2-, PO4 3- }$ and $ce{H3O+}$..
    – Adnan AL-Amleh
    Dec 2 '18 at 4:51










  • @AdnanAL-Amleh. Why don't you write all the balance equations first and post them in a question here (on MSE) ? Tell me when done and I shall have a look at the problem.
    – Claude Leibovici
    Dec 2 '18 at 5:13










  • Leibovicithanks a lot, I will post the chemical problem on (MSE), and tell you when done. Now I try to understand the derivations $ X,Y,Z,M$ as functions of $H$
    – Adnan AL-Amleh
    Dec 2 '18 at 5:35










  • @AdnanAL-Amleh. Just successive eliminations : from $(1)$, get $X$; replace in $(2)$ and get $Y$; replace in $(3)$ and get $Z$. Now, use $(5)$ to get $M$ and you are done.
    – Claude Leibovici
    Dec 2 '18 at 5:39











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1 Answer
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1 Answer
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active

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active

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active

oldest

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1














This seems to be a chemical problem. When I saw it, I suspected possible traps because of the very small numbers and I rewrote the equations as
$$a =frac{X H}{M} tag{1}$$
$$b =frac{Y H}{X} tag{2}$$
$$c =frac{Z H}{Y} tag{3}$$
$$H^2 +frac 3 {10}H = X H + 2Y H + 3ZH+dtag{4} $$
$$frac 3 {10}= M + X + Y + Ztag{5} $$



In a first step, as @user10354138 commented, I used equations $(1,2,3,5)$ to express $X,Y,Z,M$ as functions of $H$. This leads to
$$X=frac{3 a H^2}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad Y=frac{3 a b H}{10 left(a b c+a b H+a H^2+H^3right)}$$
$$Z=frac{3 a b c}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad M=frac{3 H^3}{10 left(a b c+a b H+a H^2+H^3right)}$$
Now, replace these expressions in $(4)$, reduce to same denominator (which we shall ignore assuming it is non-zero) and put everything on the same side. We shall get
$$-10 a b c d-2 a b (3 c+5 d)H+a (b (10 c-3)-10 d)H^2+10 (a b-d)H^3+(10 a+3)
H^4+10 H^5=0tag 6$$
This is just a polynomial of degree $5$ which cannot be solved analytically and for which numerical methods will be required.



Replace $a,b,c,d$ by their values; you will get very ugly values for the coefficients.



If you want to solve $(6)$ for $H$, since we know the usual ranges of $H$, use inspection with $H=10^{-pH}$. Plotting the function as a function of $pH$, you will notice a very fast decreasing function with a solution close to $pH=4$. So, start Newton iterations using $pH_0=4$. You should get the following iterates
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.10601 \
2 & 4.21036 \
3 & 4.31193 \
4 & 4.40889 \
5 & 4.49815 \
6 & 4.57469 \
7 & 4.63121 \
8 & 4.66103 \
9 & 4.66846 \
10 & 4.66886
end{array}
right)$$
that is to say $H=10^{-4.66886}=0.0000214358$.



Edit



Plotting the function given by $(6)$ is not very good. You can have a much better perception using $H=10^{-pH}$ and plot, as a function of $pH$
$$f(pH)=frac{H^2 +frac 3 {10}H } {X H + 2Y H + 3ZH+d }-1tag 7$$ and look for its zero. Zoom once around the solution to get an estimate of $4.7$.



Solving $(7)$, using as before $pH_0=4$, Newton iterations would be
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.39558 \
2 & 4.63729 \
3 & 4.66879 \
4 & 4.66886
end{array}
right)$$
which is much faster than the previous one.






share|cite|improve this answer























  • I appreciate your interest and your effort in helping for solving the math part of the chemical problem posted in Chem.Stack Exchange how do I calculate the concentration of all species present in a solution with$pu{ 0.3M}ce{ NaH2PO4}$: So the concentration of$ce{ H3PO4, H2PO4 -, HPO4 2-, PO4 3- }$ and $ce{H3O+}$..
    – Adnan AL-Amleh
    Dec 2 '18 at 4:51










  • @AdnanAL-Amleh. Why don't you write all the balance equations first and post them in a question here (on MSE) ? Tell me when done and I shall have a look at the problem.
    – Claude Leibovici
    Dec 2 '18 at 5:13










  • Leibovicithanks a lot, I will post the chemical problem on (MSE), and tell you when done. Now I try to understand the derivations $ X,Y,Z,M$ as functions of $H$
    – Adnan AL-Amleh
    Dec 2 '18 at 5:35










  • @AdnanAL-Amleh. Just successive eliminations : from $(1)$, get $X$; replace in $(2)$ and get $Y$; replace in $(3)$ and get $Z$. Now, use $(5)$ to get $M$ and you are done.
    – Claude Leibovici
    Dec 2 '18 at 5:39
















1














This seems to be a chemical problem. When I saw it, I suspected possible traps because of the very small numbers and I rewrote the equations as
$$a =frac{X H}{M} tag{1}$$
$$b =frac{Y H}{X} tag{2}$$
$$c =frac{Z H}{Y} tag{3}$$
$$H^2 +frac 3 {10}H = X H + 2Y H + 3ZH+dtag{4} $$
$$frac 3 {10}= M + X + Y + Ztag{5} $$



In a first step, as @user10354138 commented, I used equations $(1,2,3,5)$ to express $X,Y,Z,M$ as functions of $H$. This leads to
$$X=frac{3 a H^2}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad Y=frac{3 a b H}{10 left(a b c+a b H+a H^2+H^3right)}$$
$$Z=frac{3 a b c}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad M=frac{3 H^3}{10 left(a b c+a b H+a H^2+H^3right)}$$
Now, replace these expressions in $(4)$, reduce to same denominator (which we shall ignore assuming it is non-zero) and put everything on the same side. We shall get
$$-10 a b c d-2 a b (3 c+5 d)H+a (b (10 c-3)-10 d)H^2+10 (a b-d)H^3+(10 a+3)
H^4+10 H^5=0tag 6$$
This is just a polynomial of degree $5$ which cannot be solved analytically and for which numerical methods will be required.



Replace $a,b,c,d$ by their values; you will get very ugly values for the coefficients.



If you want to solve $(6)$ for $H$, since we know the usual ranges of $H$, use inspection with $H=10^{-pH}$. Plotting the function as a function of $pH$, you will notice a very fast decreasing function with a solution close to $pH=4$. So, start Newton iterations using $pH_0=4$. You should get the following iterates
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.10601 \
2 & 4.21036 \
3 & 4.31193 \
4 & 4.40889 \
5 & 4.49815 \
6 & 4.57469 \
7 & 4.63121 \
8 & 4.66103 \
9 & 4.66846 \
10 & 4.66886
end{array}
right)$$
that is to say $H=10^{-4.66886}=0.0000214358$.



Edit



Plotting the function given by $(6)$ is not very good. You can have a much better perception using $H=10^{-pH}$ and plot, as a function of $pH$
$$f(pH)=frac{H^2 +frac 3 {10}H } {X H + 2Y H + 3ZH+d }-1tag 7$$ and look for its zero. Zoom once around the solution to get an estimate of $4.7$.



Solving $(7)$, using as before $pH_0=4$, Newton iterations would be
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.39558 \
2 & 4.63729 \
3 & 4.66879 \
4 & 4.66886
end{array}
right)$$
which is much faster than the previous one.






share|cite|improve this answer























  • I appreciate your interest and your effort in helping for solving the math part of the chemical problem posted in Chem.Stack Exchange how do I calculate the concentration of all species present in a solution with$pu{ 0.3M}ce{ NaH2PO4}$: So the concentration of$ce{ H3PO4, H2PO4 -, HPO4 2-, PO4 3- }$ and $ce{H3O+}$..
    – Adnan AL-Amleh
    Dec 2 '18 at 4:51










  • @AdnanAL-Amleh. Why don't you write all the balance equations first and post them in a question here (on MSE) ? Tell me when done and I shall have a look at the problem.
    – Claude Leibovici
    Dec 2 '18 at 5:13










  • Leibovicithanks a lot, I will post the chemical problem on (MSE), and tell you when done. Now I try to understand the derivations $ X,Y,Z,M$ as functions of $H$
    – Adnan AL-Amleh
    Dec 2 '18 at 5:35










  • @AdnanAL-Amleh. Just successive eliminations : from $(1)$, get $X$; replace in $(2)$ and get $Y$; replace in $(3)$ and get $Z$. Now, use $(5)$ to get $M$ and you are done.
    – Claude Leibovici
    Dec 2 '18 at 5:39














1












1








1






This seems to be a chemical problem. When I saw it, I suspected possible traps because of the very small numbers and I rewrote the equations as
$$a =frac{X H}{M} tag{1}$$
$$b =frac{Y H}{X} tag{2}$$
$$c =frac{Z H}{Y} tag{3}$$
$$H^2 +frac 3 {10}H = X H + 2Y H + 3ZH+dtag{4} $$
$$frac 3 {10}= M + X + Y + Ztag{5} $$



In a first step, as @user10354138 commented, I used equations $(1,2,3,5)$ to express $X,Y,Z,M$ as functions of $H$. This leads to
$$X=frac{3 a H^2}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad Y=frac{3 a b H}{10 left(a b c+a b H+a H^2+H^3right)}$$
$$Z=frac{3 a b c}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad M=frac{3 H^3}{10 left(a b c+a b H+a H^2+H^3right)}$$
Now, replace these expressions in $(4)$, reduce to same denominator (which we shall ignore assuming it is non-zero) and put everything on the same side. We shall get
$$-10 a b c d-2 a b (3 c+5 d)H+a (b (10 c-3)-10 d)H^2+10 (a b-d)H^3+(10 a+3)
H^4+10 H^5=0tag 6$$
This is just a polynomial of degree $5$ which cannot be solved analytically and for which numerical methods will be required.



Replace $a,b,c,d$ by their values; you will get very ugly values for the coefficients.



If you want to solve $(6)$ for $H$, since we know the usual ranges of $H$, use inspection with $H=10^{-pH}$. Plotting the function as a function of $pH$, you will notice a very fast decreasing function with a solution close to $pH=4$. So, start Newton iterations using $pH_0=4$. You should get the following iterates
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.10601 \
2 & 4.21036 \
3 & 4.31193 \
4 & 4.40889 \
5 & 4.49815 \
6 & 4.57469 \
7 & 4.63121 \
8 & 4.66103 \
9 & 4.66846 \
10 & 4.66886
end{array}
right)$$
that is to say $H=10^{-4.66886}=0.0000214358$.



Edit



Plotting the function given by $(6)$ is not very good. You can have a much better perception using $H=10^{-pH}$ and plot, as a function of $pH$
$$f(pH)=frac{H^2 +frac 3 {10}H } {X H + 2Y H + 3ZH+d }-1tag 7$$ and look for its zero. Zoom once around the solution to get an estimate of $4.7$.



Solving $(7)$, using as before $pH_0=4$, Newton iterations would be
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.39558 \
2 & 4.63729 \
3 & 4.66879 \
4 & 4.66886
end{array}
right)$$
which is much faster than the previous one.






share|cite|improve this answer














This seems to be a chemical problem. When I saw it, I suspected possible traps because of the very small numbers and I rewrote the equations as
$$a =frac{X H}{M} tag{1}$$
$$b =frac{Y H}{X} tag{2}$$
$$c =frac{Z H}{Y} tag{3}$$
$$H^2 +frac 3 {10}H = X H + 2Y H + 3ZH+dtag{4} $$
$$frac 3 {10}= M + X + Y + Ztag{5} $$



In a first step, as @user10354138 commented, I used equations $(1,2,3,5)$ to express $X,Y,Z,M$ as functions of $H$. This leads to
$$X=frac{3 a H^2}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad Y=frac{3 a b H}{10 left(a b c+a b H+a H^2+H^3right)}$$
$$Z=frac{3 a b c}{10 left(a b c+a b H+a H^2+H^3right)}qquad qquad M=frac{3 H^3}{10 left(a b c+a b H+a H^2+H^3right)}$$
Now, replace these expressions in $(4)$, reduce to same denominator (which we shall ignore assuming it is non-zero) and put everything on the same side. We shall get
$$-10 a b c d-2 a b (3 c+5 d)H+a (b (10 c-3)-10 d)H^2+10 (a b-d)H^3+(10 a+3)
H^4+10 H^5=0tag 6$$
This is just a polynomial of degree $5$ which cannot be solved analytically and for which numerical methods will be required.



Replace $a,b,c,d$ by their values; you will get very ugly values for the coefficients.



If you want to solve $(6)$ for $H$, since we know the usual ranges of $H$, use inspection with $H=10^{-pH}$. Plotting the function as a function of $pH$, you will notice a very fast decreasing function with a solution close to $pH=4$. So, start Newton iterations using $pH_0=4$. You should get the following iterates
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.10601 \
2 & 4.21036 \
3 & 4.31193 \
4 & 4.40889 \
5 & 4.49815 \
6 & 4.57469 \
7 & 4.63121 \
8 & 4.66103 \
9 & 4.66846 \
10 & 4.66886
end{array}
right)$$
that is to say $H=10^{-4.66886}=0.0000214358$.



Edit



Plotting the function given by $(6)$ is not very good. You can have a much better perception using $H=10^{-pH}$ and plot, as a function of $pH$
$$f(pH)=frac{H^2 +frac 3 {10}H } {X H + 2Y H + 3ZH+d }-1tag 7$$ and look for its zero. Zoom once around the solution to get an estimate of $4.7$.



Solving $(7)$, using as before $pH_0=4$, Newton iterations would be
$$left(
begin{array}{cc}
n & pH_n \
0 & 4.00000 \
1 & 4.39558 \
2 & 4.63729 \
3 & 4.66879 \
4 & 4.66886
end{array}
right)$$
which is much faster than the previous one.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 3 '18 at 5:47

























answered Dec 2 '18 at 4:13









Claude Leibovici

119k1157132




119k1157132












  • I appreciate your interest and your effort in helping for solving the math part of the chemical problem posted in Chem.Stack Exchange how do I calculate the concentration of all species present in a solution with$pu{ 0.3M}ce{ NaH2PO4}$: So the concentration of$ce{ H3PO4, H2PO4 -, HPO4 2-, PO4 3- }$ and $ce{H3O+}$..
    – Adnan AL-Amleh
    Dec 2 '18 at 4:51










  • @AdnanAL-Amleh. Why don't you write all the balance equations first and post them in a question here (on MSE) ? Tell me when done and I shall have a look at the problem.
    – Claude Leibovici
    Dec 2 '18 at 5:13










  • Leibovicithanks a lot, I will post the chemical problem on (MSE), and tell you when done. Now I try to understand the derivations $ X,Y,Z,M$ as functions of $H$
    – Adnan AL-Amleh
    Dec 2 '18 at 5:35










  • @AdnanAL-Amleh. Just successive eliminations : from $(1)$, get $X$; replace in $(2)$ and get $Y$; replace in $(3)$ and get $Z$. Now, use $(5)$ to get $M$ and you are done.
    – Claude Leibovici
    Dec 2 '18 at 5:39


















  • I appreciate your interest and your effort in helping for solving the math part of the chemical problem posted in Chem.Stack Exchange how do I calculate the concentration of all species present in a solution with$pu{ 0.3M}ce{ NaH2PO4}$: So the concentration of$ce{ H3PO4, H2PO4 -, HPO4 2-, PO4 3- }$ and $ce{H3O+}$..
    – Adnan AL-Amleh
    Dec 2 '18 at 4:51










  • @AdnanAL-Amleh. Why don't you write all the balance equations first and post them in a question here (on MSE) ? Tell me when done and I shall have a look at the problem.
    – Claude Leibovici
    Dec 2 '18 at 5:13










  • Leibovicithanks a lot, I will post the chemical problem on (MSE), and tell you when done. Now I try to understand the derivations $ X,Y,Z,M$ as functions of $H$
    – Adnan AL-Amleh
    Dec 2 '18 at 5:35










  • @AdnanAL-Amleh. Just successive eliminations : from $(1)$, get $X$; replace in $(2)$ and get $Y$; replace in $(3)$ and get $Z$. Now, use $(5)$ to get $M$ and you are done.
    – Claude Leibovici
    Dec 2 '18 at 5:39
















I appreciate your interest and your effort in helping for solving the math part of the chemical problem posted in Chem.Stack Exchange how do I calculate the concentration of all species present in a solution with$pu{ 0.3M}ce{ NaH2PO4}$: So the concentration of$ce{ H3PO4, H2PO4 -, HPO4 2-, PO4 3- }$ and $ce{H3O+}$..
– Adnan AL-Amleh
Dec 2 '18 at 4:51




I appreciate your interest and your effort in helping for solving the math part of the chemical problem posted in Chem.Stack Exchange how do I calculate the concentration of all species present in a solution with$pu{ 0.3M}ce{ NaH2PO4}$: So the concentration of$ce{ H3PO4, H2PO4 -, HPO4 2-, PO4 3- }$ and $ce{H3O+}$..
– Adnan AL-Amleh
Dec 2 '18 at 4:51












@AdnanAL-Amleh. Why don't you write all the balance equations first and post them in a question here (on MSE) ? Tell me when done and I shall have a look at the problem.
– Claude Leibovici
Dec 2 '18 at 5:13




@AdnanAL-Amleh. Why don't you write all the balance equations first and post them in a question here (on MSE) ? Tell me when done and I shall have a look at the problem.
– Claude Leibovici
Dec 2 '18 at 5:13












Leibovicithanks a lot, I will post the chemical problem on (MSE), and tell you when done. Now I try to understand the derivations $ X,Y,Z,M$ as functions of $H$
– Adnan AL-Amleh
Dec 2 '18 at 5:35




Leibovicithanks a lot, I will post the chemical problem on (MSE), and tell you when done. Now I try to understand the derivations $ X,Y,Z,M$ as functions of $H$
– Adnan AL-Amleh
Dec 2 '18 at 5:35












@AdnanAL-Amleh. Just successive eliminations : from $(1)$, get $X$; replace in $(2)$ and get $Y$; replace in $(3)$ and get $Z$. Now, use $(5)$ to get $M$ and you are done.
– Claude Leibovici
Dec 2 '18 at 5:39




@AdnanAL-Amleh. Just successive eliminations : from $(1)$, get $X$; replace in $(2)$ and get $Y$; replace in $(3)$ and get $Z$. Now, use $(5)$ to get $M$ and you are done.
– Claude Leibovici
Dec 2 '18 at 5:39


















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