# 60 ml 0.2(N) H_2SO_4+40 ml 0.4(N) HCl is given. What is the pH of the solution?

Jul 6, 2017

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

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

["H"^(+)] = "0.012 + 0.016 mols H"^(+)/("60 + 40 mL") xx "1000 mL"/"1 L"

$\approx$ $\text{0.28 M}$

and consequently, an estimated $\text{pH}$ of

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

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

DISCLAIMER: No approximations are made here.

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

So, a $\text{0.2 N}$ solution of ${\text{H"_2"SO}}_{4}$ is $\text{0.1 M}$ with respect to ${\text{HSO}}_{4}^{-}$, for instance, because ${\text{H}}^{+}$ is $2 : 1$ with ${\text{H"_2"SO}}_{4}$. (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 ${H}^{+}$, we can divide by the total volume last to find the final concentration.

${\text{H"_2"SO"_4(aq) -> "HSO"_4^(-)(aq) + "H}}^{+} \left(a q\right)$

"mols HSO"_4^(-) = "0.1 M" xx "0.060 L" = color(green)("0.006 mols first H"^(+))

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

"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, ${\text{HSO}}_{4}^{-}$, and that would have the following equilibrium:

${\text{HSO"_4^(-)(aq) + "H"_2"O"(l) rightleftharpoons "SO"_4^(2-)(aq) + "H"_3"O}}^{+} \left(a q\right)$

$\text{I"" ""0.006 mols"" "-" "" "" "" ""0 mols"" "" ""0.006 mols}$
$\text{C"" "-x" "" "" "-" "" "" "+x" "" "" "" } + x$
$\text{E"" "0.006 - x" "-" "" "" "" "x" "" "" "" } 0.006 + x$

${K}_{a 2} = 1.2 \times {10}^{- 2} = \frac{\left(x\right) \left(0.006 + x\right)}{0.006 - x}$,

a non-negligible dissociation constant.

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

$x = {\text{0.00337 mols second H}}^{+}$

This means from ${\text{H"_2"SO}}_{4}$, you place

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

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

The total mols of ${\text{H}}^{+}$ 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 ${\text{H}}^{+}$ after the solution was mixed and prepared:

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

$=$ ${\text{0.2537 M H}}^{+}$ at equilibrium

Therefore, the $\text{pH}$ is:

$\textcolor{b l u e}{{\text{pH") = -log["H}}^{+}}$

$= - \log \left(\text{0.2537 M}\right)$

$= \textcolor{b l u e}{0.60}$