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.




[![enter image description here](https://i.stack.imgur.com/yNgjV.jpg)](https://i.stack.imgur.com/yNgjV.jpg)


I am posting a solution image regarding same question. 2nd hydrolysis of X2- is neglected as you did. important thing is degree of hydrolysis is not small and should not neglected.


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