Determine the [OH-] and pH of a solution that is 0.160M in F-?

Determine the [OH-] and pH of a remedy this is certainly 0.160M in F-?

F^-1 + H2O <--> HF + OH^-1
Kb for F^-1 = 1.35×10^-11
1.35×10^-11 = [HF][OH-]/[F-]
Allow X = the mol/L of F- that responds to produce X mol/L of HF and X mol/L of OH-
1.35×10^-11 = X^2 / (0.160-X)
Believe that X << 0.160 so (0.160-X) becomes just 0.160 X = 1.47x10^-6 = [OH-] pOH = 5.83 pH = 14.00-5.83 = 8.17

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