What is the pH of buffer solution containing 10 times higher molar...

Using the Henderson-Hasselbalch equation for a basic buffer:

$pOH = pK_b + \log \left( \dfrac{[Salt]}{[Base]} \right)<br><br>Given$ K_b = 10^{-5}$, we have$pK_b = -$\log$(10^{-5}) = 5$.<br><br>It is given that the concentration of the weak base is 10$ times the concentration of the salt, so $[Base] = 10 [Salt]$, which gives $\dfrac{[Salt]}{[Base]} = \dfrac{1}{10}$.

Substituting these values into the equation:

$pOH = 5 + \log \left( \dfrac{1}{10} \right)$

$pOH = 5 - 1 = 4$

At $298$ K, the relationship between pH and pOH is:

$pH + pOH = 14$

$pH = 14 - 4 = 10$

Answer: $10$