The Problem
Calculate the pH of a buffer that is $\pu{0.200M}$ $\ce{H3BO3}$ and $\pu{0.122M}$ $\ce{KH2BO3}$. The $K_\mathrm{a}$ for $\ce{H3BO3}$ is $7.3\times10^{-10}$
What I Tried
For each of these methods, I used $\pu{0.200M}$ as the concentration for the acid, $\ce{H3BO3}$, and $\pu{0.122M}$ as the concentration of its conjugate base, $\ce{H2BO3-}$.
Method one – Henderson–Hasselbalch equation
$\mathrm{p}K_\mathrm{a} = -\log(K_\mathrm{a}) = 9.13668$
${\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{0.122}{0.200}\right) = 9.13668 + (-0.21467) = 8.92201}$
Method two – Rice Table $$ \begin{array}{c|lcr} & \text{$\ce{H3BO3}$} & \text{$\ce{H3O+}$} & \text{$\ce{H2BO3-}$} \\ \hline \text{I} & 0.200 & 0 & 0.122 \\ \text{C} & -x & +x & +x \\ \text{E} & 0.200-x & x & 0.122+x \end{array} $$ use x is small approximation $K_\mathrm{a} = \ce{\frac{[\ce{H3O}][\ce{A-}]}{[\ce{HA}]}}$
$\ce{K_\mathrm{a} = \frac{0.122x}{0.200}}$
$x = 0.200 \times \frac{K_\mathrm{a}}{0.122}$
$x = 1.19672*10^{-9}$ – This equals the number of moles $\ce{H3O+}$ $\mathrm{pH} = -\log([\ce{H3O+}]) = 8.92201$
Check $x$ is small approximation: $\frac{x}{0.122} = 9.081\times10^{-9}$ which is less than $5\ \%$, so the approximation is valid.
My question
The answer I got using both of these methods, $\mathrm{pH} = 8.92$, is not correct. The only conclusion I can come to is that I have missed some nuance of this question or I am making a careless mistake. Can somebody please tell me what I'm overlooking here?