Calculate the pH of a solution that is 0.28 M in HF and 0.10M in NaF. Report Abu

Question

Calculate the pH of a solution that is 0.28 M in HF and 0.10M in NaF.
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Explanation / Answer

HF .... Ka = 7.2x10^-4 and its salt is sodium fluoride (NaF) As these formulas imply hyrdofluoric acid is a monoprotic acid. Concentrations of the substances; [HF] = 0.20 M [NaF] = 0.10 M (a) Dissociations: Weak acid ionizes only a small extent. HF(aq) H+(aq) + F-(aq) 0.20 - x M ...... x M ........ x M Ionic soluble salt ionizes completely. NaF(aq) -----> Na+(aq) + F-(aq) 0.10 M ........... 0.10 M ..... 0.10 M F- is the common ion, but the concentration of F- ion produced by the dissociation of the salt (NaF) will be very large compared to the concentration of F- ion produced by the dissociation of the acid, and hence x is neglected. Ka = [H+][F-] / [HF] This expression in some cases simplified as (since x is neglected) Ka = [H+][SALT] / [ACID] Substituting the values; 7.2x10^-4 = [H+] (0.10) / (0.20) [H+] = (7.2x10^-4 x 0.20) / 0.10 = 1.44x10^-3

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