# Question 672a6

May 3, 2017

$\text{pH} = 4.56$

You have to look up the ${K}_{b}$ of this compound's conjugate base, which is about $1.3 \times {10}^{- 6}$. Alternatively, the $\text{pKa}$ of hydrazinium is listed on the hydrazine Wikipedia as $8.10$. However, you should use the ${K}_{a}$ listed in your textbook to be more accurate.

I'll use the ${K}_{b}$ since it is more educational. Recall that:

${K}_{w} = {K}_{a} {K}_{b}$.

Therefore, you can get the ${K}_{a}$ of the acid, which is what you are actually placing into water. Remember to always use ${K}_{a}$ for acids and ${K}_{b}$ for bases.

${K}_{a} = {10}^{- 14} / \left(1.3 \times {10}^{- 6}\right) = 7.69 \times {10}^{- 9}$

The reaction itself can be denoted in general:

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

$\text{I"" ""0.10 M"" "" "" "" "" "" "-" "" "" ""0 M"" "" "" "" ""0 M}$
$\text{C"" "-x" "" "" "" "" "" "" "-" "" "+x" "" "" "" } + x$
$\text{E"" "(0.10 - x)"M"" "" "" "" "-" "" ""x M"" "" "" "" ""x M}$

And its ${K}_{a}$ can be written with a mass action expression:

${K}_{a} = {x}^{2} / \left(0.10 - x\right)$

However, since ${K}_{a} \text{<<} {10}^{- 5}$, it's a great time to make the small $x$ approximation. That is, if you're tired of using the quadratic formula.

${K}_{a} \approx {x}^{2} / \left(0.10\right)$

Therefore, we can solve for the equilibrium concentration of ${\text{H}}^{+}$ easily:

$\left[{\text{H"_3"O}}^{+}\right] = x \approx \sqrt{0.10 {K}_{a}}$

$= 2.77 \times {10}^{- 5} \text{M}$

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

$\textcolor{b l u e}{{\text{pH") = -log["H"_3"O}}^{+}}$

$= \textcolor{b l u e}{4.56}$

And another way to check whether the small $x$ approximation is appropriate is to see if the percent dissociation is less than 5%.

%"dissoc". = x/(["soln"])

= (2.77 xx 10^(-5) "M")/("0.10 M") xx 100% = 0.028% "<<" 5%#

So this approximation works extremely well. You would get pretty much the same answer if you chose to solve the quadratic equation in full.