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25.000 mL of 0.1438 M butanoic acid solution is titrated with 0.1247 M NaOH. Calculate the...

25.000 mL of 0.1438 M butanoic acid solution is titrated with 0.1247 M NaOH. Calculate the pH of titrant mixture at the following volumes of NaOH added. Ka butanoic acid = 1.52 x 10^-5

a.) 0.00 mL

b.) 16.04 mL

c.) 28.83 mL (Equivalence point)

d.) 30.30 mL

My answers were:

a.) 2.830

b.) 4.917

c.) 8.821

d.) 11.521

Expert Solution

Given,

Molarity of Butanoic Acid solution = 0.1438 M

Volume = 25 mL

=> Millimoles of Butanoic Acid = 0.1438 x 25 = 3.595

Ka = 1.52 x 10^-5

pKa = - log Ka = 4.82

a)

CH3CH2CH2COOH + H2O ------> CH3CH2CH2COO- + H3O+

0.1438 - X...............................................X.........................X

Ka = [CH3CH2CH2COO-] [H+] / [CH3CH2CH2COOH]

=> 1.52 x 10^-5 = X^2 / (0.1438 - X)

=> X = 1.48 x 10^-3 M = [H+]

pH = - log [H+] = - log (1.48 x 10^-3) = 2.83

b)

Millimoles of NaOH added = 0.1247 x 16.04 = 2

Millimoles of CH3CH2CH2COOH = 3.595

Total Volume = 25 + 16.04 = 41.04 mL

CH3CH2CH2COOH + NaOH ------> CH3CH2CH2COONa + H2O

After complete reaction,

Millimoles of CH3CH2CH2COONa (salt) produced = 2

Millimoles of CH3CH2CH2COOH(acid) remaining = 3.595 - 2 = 1.595

This solution will act as a buffer

pH of a buffer = pKa + log (Salt / Acid)

=> pH = 4.82 + log (2 / 1.595) = 4.918

c)

At equivalenve point

Millimoles of CH3CH2CH2COONa produced = 3.595

Volume of Solution = 25 + 28.83 = 53.83 mL

[CH3CH2CH2COONa] = 3.595 / 53.83 = 0.0668 M

CH3CH2CH2COONa + H2O --------> CH3CH2CH2COOH + OH- + Na+

0.0668 - X...................................................X..........................X

Kb = 10^-14 / 1.52 x 10^-5 = 6.58 x 10^-10

6.58 x 10^-10 = X^2 / (0.0668 - X)

=> X = 6.63 x 10^-6 M = [OH-]

pOH = - log [OH-] = 5.18

pH = 14 - pOH = 8.82

d)

Millimoles of NaOH added = 0.1247 x 30.30 = 3.7784

Millimoles of CH3CH2CH2COOH = 3.595

Total Volume = 25 + 30.30 = 55.30 mL

CH3CH2CH2COOH + NaOH ------> CH3CH2CH2COONa + H2O

After complete reaction,

Millimoles of CH3CH2CH2COONa (salt) produced = 3.595

Millimoles of NaOH(acid) remaining = 3.7784 - 3.595 = 0.1834

[CH3CH2CH2COONa] = 3.595 / 55.3 = 0.065 M

[NaOH] = 0.1834 / 55.3 = 0.003316 M

CH3CH2CH2COONa + H2O --------> CH3CH2CH2COOH + OH- + Na+

0.065 - X........................................................X....................0.003316+X

Kb = 6.58 x 10^-10 = (X) (0.003316+X) / (0.065 - X)

=> X = 1.29 x 10^-8 M

[OH-] = 0.003316 + X = 0.003316 M

pOH = - log [OH-] = 2.48

pH = 14 - pOH = 11.52

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