A student finds that it takes 35.0 mL of 0.20 M HCl to reach the equivalence poi

Question

A student finds that it takes 35.0 mL of 0.20 M HCl to reach the equivalence point in titrating 25 mL of an aqueous ammonia (NH3) sample. Kb for NH3 is 1.8x10–5 .

(a) What was the concentration of ammonia in the starting solution?

(b) What was the pH of the solution before addition of any acid?

(c) What would be the pH of the solution when 15.0 mL of HCl solution has been added?

(d) What would be the pH of the solution at the equivalence point?

(e) What would be the pH of the solution when 45.0 mL of HCl solution has been added?

Explanation / Answer

A) NH3 + HCl ---> NH4Cl

No of mol of HCl reacted = 35*0.2 = 7 mmol

No of mol of NH3 reacted = 7 mmol

concentration of NH3 = 7/25 = 0.28 M

b) before addition of any acid

   pkb of NH3 = -log(1.8*10^-5) = 4.7

   pH = 14 - 1/2(pkb-logC)

      = 14 - 1/2(4.7-log0.28)

      = 11.37

c)
      No of mol of NH3 taken = 7 mmol

No of mol of HCl reacted = 15*0.2 = 3 mmol

pH = 14 - (pkb+log(acid/base))

      = 14 - (4.7+log(3/(7-3))

       = 9.425

d) at the equivalence point

    concentration of salt = 7/(25+35) = 0.116 M

    pH = 7-1/2(pkb+logC)

       = 7-1/2(4.7+log0.116)

        = 5.12

e) concentration of excess HCl = (45-35)*0.2/80 = 0.025 M

     pH = -log(0.025)

         = 1.602

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