# Hydrolysis of a salt B2X

An aqueous solution of soluble salt $\ce{B2X}$ has concentration of 0.1 M. Given data :

$\ce{K_b(BOH) = 2 × 10^{–5}}$

$\ce{K_{a1}(H2X) = 2 × 10^{–7}}$

$\ce{K_{a2}(H2X) = 5 × 10^{–10}}$

What is the pH of the solution?

My attempt,

$$\ce{B+ + X2- -> BOH + HX- \,\,... (1)}$$

The hydrolysis constant of $\ce{HX-}$ comes out to be about $10^{-7}$, so I decided to neglect it (is this correct?)

$$Kh=[HX-][BOH]/[X2-][B+]$$

So the value of $$\ce{K_h=\frac{Kw}{K_{a2}*K_b}}$$

Upon substituting the values I get the Kh=1..(something i have never seen before)

So this means that for the reaction

$$\ce {B+ + X2- -> BOH + HX-}$$

We will end up with 0.05M of $\ce{B+}$ , 0.05M of $\ce{of X2-}$, 0.05M of $\ce{BOH}$ , and 0.05M $\ce{HX-}$ if we start with 0.1 M of $\ce{B+}$ and 0.1 M of $\ce{X2-}$

Now, I do know that

$$\ce{pH=7+\frac{pK_a + pK_b}{2}}$$

I also know that this is true when the hydrolysis constant is smaller than 1. Therefore I cannot use if for this problem.

(Nevertheless, I disregarded this and just used it. I got the answer to be 9.305)

So I am stuck over here.

I got another idea.

When I now look at the sum, the resulting solution looks like a mixture of two buffer solutions,

Buffer 1: $\ce{BOH/B+}$

Buffer 2: $\ce{HX-/H2X}$

I don't know whether this observation is correct. Even if it is correct, I am not sure whether it be used to solve the problem.

• Hello Vedant, welcome to chemistry SE. It isn't clear as to what you are asking. It appears that you have to determine the the final pH. However, I am not sure. – CoffeeIsLife May 19 '17 at 2:00
• I am sorry!...yes the final pH is what I am asking. I have edited the question(added what I was asking for) – Vedant Parwal May 19 '17 at 2:26 