you will be working originally with 40.0 mL of a 0.27 M solution of HOCl ( K a =

Question

you will be working originally with 40.0 mL of a 0.27 M solution of HOCl ( Ka= 3.5 x 10-8 )

1) Determine the pH of your solution

2) Add 50.0 mL of 0.20 M NaOCl to your solution. Find the pH

3) Add 7.0 mL of 0.040 M HBr to the solution from task #3. Find the pH.

4) In a titration, 7.0 mL of 0.50 M KOH is added to your original solution. Find the pH.

5) What volume of 0.50 M KOH must be added to the original solution to produce a solution which would act as a buffer more than any other time in the titration (half-neutralization point? i think)? What would the pH of the solution be at that point?

6) What volume of 0.50 M KOH must be added to the original solution to reach the equivalence point in the titration? What would the pH of the solution be at that point?

7) If a volume of 27.0 mL of 0.50 M KOH is added to the original solution in a titration, what would the pH of the solution be at that point?

If a strong acid was used in place of the weak acid in the titration referenced in the last 4 tasks (same volume, same concentration) name two noteworthy and appropriate similarites in the process and two noteworthy and appropriate differences.

Explanation / Answer

1) HOCl <--------> H+ + OCl- , [HOCl] at equi = ( 0.04x0.27-x)/0.04, [H+]=[OCl-] = x/0.04,

Ka = [H+][OCl-]/[HOCl] = 3.5 x10^ -8 = x^2/( 0.0108-x)(0.04,

solving we get x = H+ moles = 3.88 x 10^ -6 , [H+] = 3.88 x10^ -6/0.04 = 9.7 x 10^ -5,

pH = -log( 9.7 x10^ -5) = 4.013,

2) pH = pka + log[NaOCl]/[HOCl] , [NaOCl] = ( 0.2 x0.05)/(0.05+0.04) =0.111,

[HOCl] =( 0.04x0.27)/(0.05+0.04) = 0.12 , pka = -log(3.5x10^ -8) = 7.456

pH = -log(3.5 x10^ -8)+ log( 0.11/0.12) = 7.418,

3) after addition of HBr [NaOCl] = ( 0.11x0.09-0.007x0.04)/(0.09+0.007) = 0.09917,

[HOCl] = ( 0.12 x0.09+0.007x0.04)/(0.09+0.007) = 0.114,

pH = 7.456+log(0.09917/0.114) = 7.395,

4) after KOH addition [KOCl] = ( 0.007x0.5)/(0.007+0.04) = 0.0744,

[HOCl] = ( 0.04x0.27-0.007x0.5) /(0.007+0.04) = 0.1553,

pH = 7.456+ log( 0.0744/( 0.1553) = 7.136,

5) at half neutralisation point pH = pka= 7.456 , means [KOCl]=[HOCl],

let V be vol , then [KOCl] = 0.5V/(V+0.04) = [HOCl]=(0.04x0.27-0.5V)/(V+0.04),

0.5V = (0.0108-0.5V) , V = 0.0108 liters = 10.8 ml ,

6) at equivalence point moles of KOH= moles of HOCl , they react to give same moles of KOCl

KOH + HOCl -----> OCl- + K+ , hence 0.5V = 0.04x0.27, V = 0.0216 liter = 21.6 ml

OCl- + H2O --> HOCl + OH- , total vol = ( 0.0216+0.04) =

Kb for KOCl = Kw/Ka = 10^ -14/(3.5 x10^ -8) = 2.857 x10^ -7 ,

Kb = 2.857 x10^ -7 = x^2/( 0.0108)(0.0616),

x = 1.37 x10^ -5 = OH_ moles , [OH-] =( 1.37x10^ -5)/(0.0616) = 2.224 x10^ -4 ,

pOH = -log( 2.224 x10^ -4) = 3.65, pH = 10.347

7) excess OH- after neutralisation = ( 0.027x0.5-0.04x0.27) = 0.0027,

[OH-] =0.027/ (0.027+0.04) = 0.403,

pOH = -log( 0.403) = 0.3947, pH = 13.6 ,

If strong acid is used instead of weak acids then

pH after excess addition of KOH will be similar like that of weak acid

however it differs in othres

pH initially , pH at half equivalence point and at equivalence point, strong acid and weak acids shows differences

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