Boric acid behaves as a monobasic acid, B(OH)₃ + 2H₂O ⇌ [B(OH)₄]⁻ + H₃O⁺, with pKa = 9.24…

Step 1. Hydrogen ions. $[\text{H}^+] = \sqrt{5.75 \times 10^{-10} \times 0.050} = 5.36 \times 10^{-6}\ \text{mol/L}$.

Step 2. pH. $\text{pH} = -\log_{10}(5.36 \times 10^{-6}) = 5.27$.

Step 3. Percentage reacted. The borate ion concentration equals $[\text{H}^+]$, so the percentage is $\dfrac{5.36 \times 10^{-6}}{0.050} \times 100 = 0.0107\%$.

Step 4. The buffer. The acid is $100.0 \times 0.100 = 10.0\ \text{mmol}$ and the alkali is $25.0 \times 0.100 = 2.50\ \text{mmol}$. The alkali changes $2.50\ \text{mmol}$ of acid into borate, so $7.50\ \text{mmol}$ of acid and $2.50\ \text{mmol}$ of borate remain.

Step 5. Henderson-Hasselbalch. $\text{pH} = 9.24 + \log_{10}(2.50 / 7.50) = 9.24 - 0.477 = 8.76$.