Let [H+] = [As-] = x, and [HAs] = 0.352 - x. Initially assume that x<< 0.352. Then:
3.0X10^-4 = x^2 / 0.352
x = 0.029 = [H+]
Now, since x is about 8% of 0.352, you really should go back and keep the x value in the denominator. So:
3.0X10^-4 = x^2 / (0.352 - x)
Solving this will require rearranging this equation into the form ax^2 + bx + c = 0, and using the quadratic formula to solve for x. Once you find, x, you defined that to be [H+]. Find the pH from that value.
Answers & Comments
_________ HC₉H₇O₄(aq) + H₂O(l) ⇌ C₉H₇O₄⁻(aq) + H₃O⁺(aq) …. Ka = 3.0 × 10⁻⁴
Initial: _____ 0.352 M _____________ 0 M _______ 0 M
Change: _____ -y M _____________ +y M ______ +y M
At eqm: __ (0.352 - y) M ___________ y M _______ y M
As Ka is small, assume that 0.352 ≫ y, i.e. (0.352 - y) M ≈ 0.352 M
Ka = [C₉H₇O₄⁻] [H₃O⁺] / [HC₉H₇O₄]
y² / 0.352 = 3.0 × 10⁻⁴
y = 0.010
pH = -log[H₃O⁺] = -log(0.010) = 2
Equilibrium: HAs <--> H+ + As-
Ka = [H+][As-]/[HAs] = 3.0X10^-4
Let [H+] = [As-] = x, and [HAs] = 0.352 - x. Initially assume that x<< 0.352. Then:
3.0X10^-4 = x^2 / 0.352
x = 0.029 = [H+]
Now, since x is about 8% of 0.352, you really should go back and keep the x value in the denominator. So:
3.0X10^-4 = x^2 / (0.352 - x)
Solving this will require rearranging this equation into the form ax^2 + bx + c = 0, and using the quadratic formula to solve for x. Once you find, x, you defined that to be [H+]. Find the pH from that value.