A 38.2 mL sample of a 0.438 M aqueous hypochlorous acid solution is titrated wit
ID: 991402 • Letter: A
Question
A 38.2 mL sample of a 0.438 M aqueous hypochlorous acid solution is titrated with a 0.346 M aqueous potassium hydroxide solution. What is the pH after 26.6 mL of base have been added?
When a 28.6 mL sample of a 0.450 M aqueous hydrocyanic acidsolution is titrated with a 0.328 M aqueous barium hydroxide solution, what is the pH at the midpoint in the titration?
What is the pH at the equivalence point in the titration of a 16.1 mL sample of a 0.364 M aqueous hydrofluoric acid solution with a 0.484M aqueous sodium hydroxide solution?
Explanation / Answer
a)
A 38.2 mL sample of a 0.438 M aqueous hypochlorous acid solution is titrated with a 0.346 M aqueous potassium hydroxide solution. when 26.6 mL of KOH is added. 26.6x0.346=9.2036 mmol of KClO is formed and 9.2036 mole of HClO gets consumed by base.
so reaction mixture have 9.2036 mmol of KClO and (38.2x0.438)- 9.2036 =16.7316-9.2036=7.528
pH of this reaction mixture is given by pH= pKa + log([ KClO]/[HClO ]) ;
pH= 7.40 + log([ KClO]/[HClO ])= 7.40 + log(9.2036/7.528)= 7.40 +0.08728 = 7.48728 ~7.49
b) Similary pH= pkHCN + Log [(CN-)]/[HCN] ; pH= 9.31 + Log [(CN-)]/[HCN] ; At mid point of tritation half of [HCN] is consumed by base. thus[(CN-)]/[HCN] =1. so pH=pKHCN =9.31
C) At equivalence point , there will be 16.1*0.364=5.8604 mmol of NaF.
volume of NaOH required= 16.1 x 0.364 / 0.484= 12.11 mL
so [NaF]= 5.8604/ (12.11+16.1)= 5.8604/28.21 = 0.2077 M
F- + H2O---> HF + OH- ; Kb=[HF][OH]/[F-] ; Kh= [HF][OH-][H+]/ [H+][F-] ; Kb= Kw/Ka= 1.47 x 10–11
0.2077-x -----> x x
So 1.47x10-11= x²/(0.2077-x) assume that 0.2077 >> x ; so 1.47x10-11= x²/(0.2077) or x2 = 0.2077*1.47x10-11
x=1.7474*10-6 M
So pOH= -log (1.7474*10-6) =5.7576 so pH=14 - 5.7576 =8.2424