Please help me with this problem

Question

4. Calculate the pH of 0.90M HPO4 2- 

​Ka HPO4 2-  = 4.5 x 10-13  

(a) 7 

(b) 14 

(c) 12.3 

(d) 8.20 

(e) 6.20 

Answer 1

K_A = [HPO₄^2-][H⁺]/[HPO₄^-] let x be the H⁺ concentration from movement to equilibrium:

K_A = 4.5 x 10^-13 = x*(10^-7+x)/(.9-x) which results in a quadratic. It is safe to assume x <<.9:

4.5 x 10^-13 = x*(10^-7+x)/.9 It looks like they also assume x >> 10^-7 (not so great in my opinion)

in which case the equation becomes

4.5 x 10^-13 = x²/.9 solve for x, then take -logx