# What is the "pH" for a solution containing "0.500 M" of a weak triprotic acid whose first K_a is 5.78 xx 10^(-10)? Assume only the first proton is significant. Report your answer to two significant figures.

Mar 4, 2017

The ICE table is the typical one for monoprotic weak acids, since we only look at the first proton here:

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

$\text{I"" "0.500" "" "" "-" "" "" "0" "" "" "" "" } 0$
$\text{C"" "-x" "" "" "-" "" "+x" "" "" "" } + x$
$\text{E"" "0.500-x" "-" "" "" "x" "" "" "" "" } x$

The mass action expression is:

${K}_{a 1} = \left(\left[\text{H"_2"A"^(-)]["H"_3"O"^(+)])/(["H"_3"A}\right]\right) = 5.78 \times {10}^{- 10}$

$= {x}^{2} / \left(0.500 - x\right)$

This ${K}_{a 1}$ is small, so to a great approximation, $x = \left[{\text{H}}^{+}\right]$ is indeed small in comparison to $\text{0.500 M}$. Thus, we make the approximation so that

$5.78 \times {10}^{- 10} \approx {x}^{2} / 0.500$

and so:

["H"^(+)] = sqrt(5.78 xx 10^(-10) cdot 0.500) " M" = 1.7 xx 10^(-5) $\text{M}$

and the $\text{pH}$ would just be:

"pH" = -log["H"^(+)] = 4.769

To two sig figs it would then be:

$\textcolor{b l u e}{\text{pH} = 4.8}$