What will be the pH of a buffer solution having an equal concentration of B⁻ and HB (Kᵦ = 10⁻¹⁰ for B⁻)

The correct answer is 4.

To understand why this is the correct answer, we can analyze the components of the buffer solution and use the Henderson-Hasselbalch equation.

First, let's identify the components of the buffer system:
- $\mathrm{HB}$ is a weak acid.
- $B^{-}$ is its conjugate base.

We are given the base dissociation constant ($K_b$) for the conjugate base $B^{-}$:
$K_b={10}^{-10}$

We can find the acid dissociation constant ($K_a$) for the conjugate acid $\mathrm{HB}$ using the relation between $K_a$, $K_b$, and the ion-product constant of water ($K_w={10}^{-14}$ at 25 °C):
$K_a\times K_b=K_w$

Substituting the values, we get:
$K_a\times {10}^{-10}={10}^{-14}$

Solving for $K_a$:
$K_a=\frac{{10}^{-14}}{{10}^{-10}}={10}^{-4}$

Next, we calculate the $p{K}_a$ of the weak acid $\mathrm{HB}$:
$p{K}_a=-{log}_{10}\left(K_a\right)$

Substituting $K_a={10}^{-4}$:
$p{K}_a=-{log}_{10}\left({10}^{-4}\right)=4$

The Henderson-Hasselbalch equation for an acidic buffer is:
$\mathrm{pH}=p{K}_a+{log}_{10}\left(\frac{\left[B^{-}\right]}{\left[\mathrm{HB}\right]}\right)$

The problem states that the buffer solution has equal concentrations of $B^{-}$ and $\mathrm{HB}$:
$\left[B^{-}\right]=\left[\mathrm{HB}\right]$

Therefore, the ratio of their concentrations is:
$\frac{\left[B^{-}\right]}{\left[\mathrm{HB}\right]}=1$

Substituting this back into the Henderson-Hasselbalch equation:
$\mathrm{pH}=p{K}_a+{log}_{10}\left(1\right)$

Since the logarithm of 1 to any base is 0 (${log}_{10}\left(1\right)=0$), we get:
$\mathrm{pH}=p{K}_a+0$

Thus:
$\mathrm{pH}=p{K}_a=4$