A solution with a pH of 8.73 is prepared by adding water to 0.614 mol of NaX to

Question

A solution with a pH of 8.73 is prepared by adding water to 0.614 mol of NaX to make 2.50 L of solution. What is the pH of the solution after 0.219 mol of HX is added?

Explanation / Answer

Exact answer for your question : You need to find the Ka of HX The concentration of NaX is 0.614/2.5 =0.246 M since HX is a weak acid the salt will hydrolyze according to .. .. .. .. .. .. X- +H2O HX + OH- Initial .. .. 0.246 React .. .. ..y Produce .. .. .. .. .. .. .. .. .. y .. .. .. y At equil. 0.246-y .. .. .. .. .. y .. .. .. y Kb= [HX][OH-]/[X-] = y^2/(0.246-y) (1) but pH=8.73 => pOH =14-8.73 =5.27 => [OH-]=10^-5.27 =y. Substitute in (1) and you get Kb =(10^-5.27)^2) / (0.246-(10^-5.27)) =1.17*10^-10 But Kb=Kw/Ka => Ka=Kw/Kb = (10^-14)/(1.17*10^-10) = 8.55*10^-5 (and thus pKa= -log(8.55*10^-5) = 4.07) Now let's go to the problem after adding the acid.Use the Henderson-Hasselbach equation pH=pKa+log[X-]/[HX] since we assume no change in volume, the ratio of concentratons is equal to the ratio of moles (volume is simplified) thus we get pH= 4.07+log(0.614/0.219) = 4.52

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