A 25.0 mL sample of 0.125 molL1 pyridine ( K b=1.7×109) is titrated with 0.100 m

Question

A 25.0 mL sample of 0.125 molL1 pyridine (Kb=1.7×109) is titrated with 0.100 molL1HCl.

Part A

Calculate the pH at 0 mL of added acid.

Express your answer using two decimal places.

Part B

Calculate the pH at 10 mL of added acid.

Express your answer using two decimal places.

Part C

Calculate the pH at 20 mL of added acid.

Express your answer using two decimal places.

Part D

Calculate the pH at equivalence point.

Express your answer using two decimal places.

Part E

Calculate the pH at one-half equivalence point.

Express your answer using two decimal places.

Part F

Calculate the pH at 40 mL of added acid.

Express your answer using two decimal places.

Part G

Calculate the pH at 50 mL of added acid.

Express your answer using two decimal places.

Explanation / Answer

A 25.0 mL sample of 0.125 molL1 pyridine (Kb=1.7×109) is titrated with 0.100 molL1HCl.

a)

initially

Kb = [HB+][OH-]/[B]

1.7*10^-9 = x*x/(0.125-x)

x = 1.457*10^-5

[OH-] = 1.457*10^-5

pOH = -log(1.457*10^-5) = 4.8365

pH = 14-4.8365 = 9.1635

b)

after 10 ml of acid

mmol of H+ = MV = 10*0.1 = 1 mmol

mmol of base = 0.125*25 = 3.125 mmol

mmol of base left = 3.125-1 = 2.125

mmol of conjugate = 0+ 1

this is a buffer

so

pOH = pKb + log(HB+/B)

pOH = -log(1.7*10^-9) + log(1/2.125) = 8.4421

pH = 14-8.4421 = 5.5579

c)

V = 20 mL

mmol of H+ = MV = 20*0.1 = 2 mmol

mmol of base = 0.125*25 = 3.125 mmol

mmol of base left = 3.125-2 = 1.125

mmol of conjugate = 0+ 2

this is a buffer

so

pOH = pKb + log(HB+/B)

pOH = -log(1.7*10^-9) + log(2/1.125) = 9.01942

pH = 14-9.01942= 4.98058

d)

in equivalence point

mmol of base = MV = 25*0.125 = 3.125

volume of acid = mmol/M = 3.125/0.1 = 31.25 mL

totla V= 31.25+25 = 56.25 mL

so..

[HB+]= m1*v1 = 3.125 /56.25 = 0.05555

Ka = [H+][B]/[HB+]

KA = Kw/Ka = (10^-14)/(1.7*10^-9) = 5.882*10^-6

5.882*10^-6 = x*x/(0.05555-x)

x = 5.65*10^-4

pH = -log( 5.65*10^-4

pH= 3.2479

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