# 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.

##### 1 Answer
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:

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