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We’ve two solutions :

Solution 1 $\ce{HCOOH}$ its concentration is $c_1=10^{-2}\ \mathrm{mol/l}$ and its volume is $v_1 = 50\ \mathrm{ml}$ and its $\mathrm{pH}_1 = 2.9$

Solution 2 $\ce{CH_3COOH}$ its concentration is $c_2=10^{-2}\ \mathrm{mol/l}$ and its volume is $v_2 = 50\ \mathrm{ml}$ and its $\mathrm{pH}_2 = 3.4$

How would be the equation and the ice table, and what is the $\mathrm{pH}$ of the mixture of these two solutions

I used number just to understand how that is work, no other reason.

I tried to do $\dfrac {\mathrm{pH}_1+\mathrm{pH}_2}{2}$, but it’s trivial.

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  • $\begingroup$ Welcome to Chemistry Stack Exchange! Please add what you have attempted towards solving the problem into the body of your question. For more information, see the site's homework policy for how to ask homework questions. Thanks! $\endgroup$ – jonsca Apr 25 '14 at 21:23
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Given the following

$[CH_3COOH]=1 \cdot 10^{-2}\,mol\,L^{-1}$

$[HCOOH]=1 \cdot 10^{-2}\,mol\,L^{-1}$

$K_{z-CH_3COOH}=1.8 \cdot 10^{-5}\,mol\,L^{-1}$

$K_{z-HCOOH}=1.8 \cdot 10^{-4}\,mol\,L^{-1}$

$[H_3O^{+}]_1 = 1.3 \cdot 10^{-3}\,mol\,L^{-1}$

$[H_3O^{+}]_2 = 4.0 \cdot 10^{-4}\,mol\,L^{-1}$

Seeing that we could write

$HCOOH + CH_3COOH + 2H_2O\rightarrow HCOO^{-} + CH_3COO^{-}+2H_3O^{+}$

We can derive a equilibrium

$\frac{[HCOO^{-}][CH_3COO^{-}][H_3O^{+}]^{2}}{[HCOOH][CH_3COOH]}=\frac{[HCOO^{-}][H_3O^{+}]}{[HCOOH]}\frac{[CH_3COO^{-}][H_3O^{+}]}{[CH_3COOH]}=K_{z-hcooh}K_{z-ch_3cooh}$

I think we still need to know the original concentrations of $CH_3COOH$ and $HCOOH$, we can calculate

$[CH_3COO^{-}]=\frac{[CH_3COOH]}{[H_3O^{+}]_2}K_{z-CH_3COOH}= 4.5000 \cdot 10^{-4}\,mol\,L^{-1}$

for $[HCOO^{-}]$ we get $1.3846 \cdot 10^{-3}\,mol\,L^{-1}$ (keeping some significance). So...

$n_{HCOOH}=1.1385 \cdot 10^{-2}\,mol\,L^{-1}$ (total)

$n_{CH_ 3COOH}=5.225 \cdot 10^{-3}\,mol\,L^{-1}$ (total)

Solving for $y$ (neglecting the already present $H_3O^{+}$ of water)

$K_{z-HCOOH}=\frac{[HCOO^{-}][H_3O^{+}]}{[HCOOH]}=\frac{x(x+y)}{(0.5n_{HCOOH}-x)}$

gives $y=-0.00018-x+\frac{1.02465 \cdot 10^{-6}}{x}$, using $y$ in

$K_{z-CH_3COOH}=\frac{[CH_3COO^{-}][H_3O^{+}]}{[CH_3COOH]}=\frac{y(x+y)}{(0.5n_{CH_3COOH}-y)}$

gives $x = 0.000884322 \vee x = 0.00690291$ (there are more solutions, but these are the only real positive ones). From this we can calculate $y=0.0000943623 \vee y = -0.00693447$. $y$ should be positive, thus $x = 0.000884322$ and $y=0.0000943623$. This gives us a $pH$ of

$pH=-\log(0.000884322+0.0000943623)\approx 3.01$

Which seems to me as correct (it will be slightly more acidic than when adding the amounts of $H^{+}$).

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  • 1
    $\begingroup$ Checking an online pH calculator, your answer seems correct. You have the right idea of combining both equilibria in a single calculation, though your solution seems a little more complex to read (it could be just me though!). There's a nice,organized and easily extendable way to solve any type of acid-base pH problem, based simply on charge and matter balance equations. If you're curious, check out some of my previous answers to other problems, e.g. here and here. $\endgroup$ – Nicolau Saker Neto May 28 '14 at 0:25
  • $\begingroup$ Another source with the mathematical description of these solutions is given here. $\endgroup$ – Nicolau Saker Neto May 28 '14 at 0:32
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    $\begingroup$ @NicolauSakerNeto That's good to hear. This is actually more simple than it looks like. You can extend this by solving more equations with more unknown. In this example we used x, y, but if I would be including $H_2O$ we would get x, y, x and an extra equation $K_w=[OH^{-}][H_3O^{+}]$. Thanks for the links, I'll look after it. $\endgroup$ – Jori May 28 '14 at 7:39
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    $\begingroup$ Also, it seems the links provided do essentially the same thing as me. Setting up equations and then substituting. $\endgroup$ – Jori May 28 '14 at 7:42
  • $\begingroup$ @NicolauSakerNeto Can you give the link of your online pH calculator? $\endgroup$ – Jori Jul 2 '14 at 22:33
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$-\log[H]=\text{p(H)}$ you can calculate concentration of $[H]$ for each acid, then add the $[H]$ for two acids and calculate $\text{p(H)}$ by $-\log[H]$.

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  • $\begingroup$ I can't understand what are you trying to do, by the way $pH=-\log [H^+]$ doesn't works for weak acids. $\endgroup$ – Hedwig Apr 27 '14 at 12:33
  • $\begingroup$ pH=−log[H+] work for all acids.you have pH for tow acids then you calculate concentration [H+] by pH=-log[H+] and addition [H+] for tow acids then calculate pH by pH=-log[H+] $\endgroup$ – mansour Apr 28 '14 at 13:38

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