(1) A 40.0 mL sample of 0.134 M HNO2 is titrated with 0.262 M KOH. (Ka for HNO2

Question

(1) A 40.0 mL sample of 0.134 M HNO2 is titrated with 0.262 M KOH. (Ka for HNO2 is 4.57×104.) Determine the pH at the equivalence point for the titration of HNO2 and KOH

(2) Consider the titration of a 25.0-mL sample of 0.110 M HC2H3O2 with 0.130 M NaOH. (The value of Ka for HC2H3O2 is 1.8×105.) Determine the pH after adding 4.0 mL of base beyond the equivalence point.

(3) Consider the titration of a 25.0-mL sample of 0.175 M CH3NH2 with 0.145 M HBr. (The value of Kb for CH3NH2 is 4.4×104.) (a) Determine the pH at one-half of the equivalence point. (b) Determine the pH after adding 5.0 mL of acid beyond the equivalence point.

Explanation / Answer

1)
at equivalence point salt is formed with weak acid and strong base then,
pH = 7 + 1/2 (pKa + logC)
For C we have to know the volume of KOH then,
M1V1 = M2V2
0.134*40 = 0.262*V2
V2 = 20.458 mL
pH = 7 + 1/2 (-log(4.57*10^-4) + log(20.458*0.262/(40+20.458))
pH = 8.144
2)
volume of NaOH V2 = M1V1/M2 = 0.11*25/0.13 = 21.154
concentration beyond equivalence point = 4/(25+21.153+4)*0.13 = 1.0368*10^-2
pOH = -log(OH-)
pOH = -log(1.0368*10^-2)
pOH = 1.9843
pH = 14 - 1.9843 = 12.0157
3)
volume of HBr V2 = M1V1/M2 = 0.175*25/0.145 = 30.172
at half equivalence point pH = pKa
pH = -log(4.4*10^-4)
pH = 3.356
concentration beyond equivalence point = 5/(31.172+25+5)*0.145 = 1.1852*10^-2
pH = -log(1.1852*10^-2)
pH = 1.9262

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