$60 m l$ $60 ml$ $0.2 (N) H_{2} S O_{4} + 40 m l$ $0.2(N) H_2SO_4+40 ml$ $0.4 (N) H C l$ $0.4(N) HCl$ is given. What is the pH of the solution?

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$60 m l$ $60 ml$ $0.2 (N) H_{2} S O_{4} + 40 m l$ $0.2(N) H_2SO_4+40 ml$ $0.4 (N) H C l$ $0.4(N) HCl$ is given. What is the pH of the solution?

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Answer 1 · 2017-07-06T02:05:44

I got $pH = 0.60$ $"pH" = 0.60$ . Notice how you can't really ignore the fact that the second proton on $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ will dissociate off of a weak acid.

If you assume that sulfuric acid dissociates 100% twice, you would get a total of $0.006 mols \times 2 = {0.012 mols H}^{+}$ $"0.006 mols" xx 2 = "0.012 mols H"^(+)$ from $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ , which would give you a total $H^{+}$ $"H"^(+)$ concentration of

$[H^{+}] = \frac{{0.012 + 0.016 mols H}^{+}}{60 + 40 mL} \times \frac{1000 mL}{1 L}$ $["H"^(+)] = "0.012 + 0.016 mols H"^(+)/("60 + 40 mL") xx "1000 mL"/"1 L"$

$\approx$ $~~$ $0.28 M$ $"0.28 M"$

and consequently, an estimated $pH$ $"pH"$ of

$pH \approx - log (0.28 M) = 0.55$ $"pH" ~~ -log("0.28 M") = 0.55$ ,

which is about $7.7 %$ $7.7%$ error. If your professor is OK with less than $5 %$ $5%$ accuracy error, then $0.55$ $0.55$ is not quite good enough...

DISCLAIMER: No approximations are made here.

Normality is defined with respect to the $H^{+}$ $"H"^(+)$ or ${OH}^{-}$ $"OH"^(-)$ the solute dissociates into the solution.

So, a $0.2 N$ $"0.2 N"$ solution of $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ is $0.1 M$ $"0.1 M"$ with respect to ${HSO}_{4}^{-}$ $"HSO"_4^(-)$ , for instance, because $H^{+}$ $"H"^(+)$ is $2 : 1$ $2:1$ with $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ . (The normality definition assumes 100% dissociation of all protons.)

Note that both sulfuric acid and hydrochloric acid are strong acids, which in principle have a 100% dissociation of their first proton.

If we calculate the mols of $H^{+}$ $H^(+)$ , we can divide by the total volume last to find the final concentration.

$H_{2} {SO}_{4} (a q) \to {HSO}_{4}^{-} (a q) + H^{+} (a q)$ $"H"_2"SO"_4(aq) -> "HSO"_4^(-)(aq) + "H"^(+)(aq)$

${mols HSO}_{4}^{-} = 0.1 M \times 0.060 L = {0.006 mols first H}^{+}$ $"mols HSO"_4^(-) = "0.1 M" xx "0.060 L" = color(green)("0.006 mols first H"^(+))$

$HCl (a q) \to H^{+} (a q) + {Cl}^{-} (a q)$ $"HCl"(aq) -> "H"^(+)(aq) + "Cl"^(-)(aq)$

${mols H}^{+} = 0.4 M \times 0.040 L = {0.016 mols H}^{+}$ $"mols H"^(+) = "0.4 M" xx "0.040 L" = color(green)("0.016 mols H"^(+))$

However, the second sulfuric acid proton is another story.

If we keep going without any approximations, we see that sulfuric acid dissociates into the weak acid, ${HSO}_{4}^{-}$ $"HSO"_4^(-)$ , and that would have the following equilibrium:

${HSO}_{4}^{-} (a q) + H_{2} O (l) ⇌ {SO}_{4}^{2 -} (a q) + H_{3} O^{+} (a q)$ $"HSO"_4^(-)(aq) + "H"_2"O"(l) rightleftharpoons "SO"_4^(2-)(aq) + "H"_3"O"^(+)(aq)$

$I 0.006 mols - 0 mols 0.006 mols$ $"I"" ""0.006 mols"" "-" "" "" "" ""0 mols"" "" ""0.006 mols"$
$C - x - + x + x$ $"C"" "-x" "" "" "-" "" "" "+x" "" "" "" "+x$
$E 0.006 - x - x 0.006 + x$ $"E"" "0.006 - x" "-" "" "" "" "x" "" "" "" "0.006 + x$

$K_{a 2} = 1.2 \times 10^{- 2} = \frac{(x) (0.006 + x)}{0.006 - x}$ $K_(a2) = 1.2 xx 10^(-2) = ((x)(0.006 + x))/(0.006 - x)$ ,

a non-negligible dissociation constant.

Solving this via the quadratic formula gives a physically reasonable $x$ $x$ as

$x = {0.00337 mols second H}^{+}$ $x = "0.00337 mols second H"^(+)$

This means from $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ , you place

${0.006 mols first H}^{+} + {0.00337 mols second H}^{+}$ $"0.006 mols first H"^(+) + "0.00337 mols second H"^(+)$

$= {0.009 mols H}^{+}$ $= "0.009 mols H"^(+)$ into solution (to three decimal places).

The total mols of $H^{+}$ $"H"^(+)$ in solution is then:

$overbrace("0.00937 mols H"^(+))^("H"_2"SO"_4) + overbrace("0.016 mols H"^(+))^("HCl") = "0.025 mols H"^(+)$

And by using the total volume of the solution, we then get the new concentration of $"H"^(+)$ after the solution was mixed and prepared:

$"0.02537 mols H"^(+)/("60 + 40 mL") xx "1000 mL"/"1 L"$

$=$ $"0.2537 M H"^(+)$ at equilibrium

Therefore, the $"pH"$ is:

$color(blue)("pH") = -log["H"^(+)]$

$= -log("0.2537 M")$

$= color(blue)(0.60)$

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$60 m l$ $60 ml$ $0.2 (N) H_{2} S O_{4} + 40 m l$ $0.2(N) H_2SO_4+40 ml$ $0.4 (N) H C l$ $0.4(N) HCl$ is given. What is the pH of the solution?

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$60 m l$ $60 ml$ $0.2 (N) H_{2} S O_{4} + 40 m l$ $0.2(N) H_2SO_4+40 ml$ $0.4 (N) H C l$ $0.4(N) HCl$ is given. What is the pH of the solution?