Question

A 25.0 mL sample of 0.125 M acetic acid is titrated with 0.125 M NaOH. Calculate the pHs after each of the following volumes of base have been added: 12.5 mL 15 mL

Answer 1

1.

25 mL of 0.125 M CH3COOH

Molarity = Moles / Liter

So, moles of CH3COOH = 0.125 M x 0.025 L = 0.003125 moles

12.5 mL of 0.125 M NaOH

Moles of NaOH = 0.125 M x 0.0125 L = 0.0015625 moles

So, 0.0015625 moles of NaOH will neutralize 0.0015625 moles of CH3COOH. So, excess of CH3COOH left

= 0.003125 moles - 0.0015625 moles = 0.0015625 moles

Total volume = 25 mL + 12.5 mL = 37.5 mL = 0.0375 L

So, [CH3COOH] = 0.0015625 moles / 0.0375 L = 0.0417 M

Now, CH₃COOH $\rightleftharpoons$ H⁺ + CH₃COO^-

Ka = [H⁺] [CH₃COO^-] / [CH₃COOH]

$\Rightarrow$ 1.8 x 10^-5 = (x) (x) / (0.0417 - x)

$\Rightarrow$ 1.8 x 10^-5 = x² / 0.0417 Neglecting x term in denominator, since Ka value is very small

$\Rightarrow$ x² = 7.506 x 10^-7

$\Rightarrow$ x= 8.66 x 10^-4

So, [H⁺] = x= 8.66 x 10^-4  

pH = - log  [H⁺] = - log (8.66 x 10^-4) = 3.06

2.

25 mL of 0.125 M CH3COOH

Molarity = Moles / Liter

So, moles of CH3COOH = 0.125 M x 0.025 L = 0.003125 moles

15 mL of 0.125 M NaOH

Moles of NaOH = 0.15 M x 0.0125 L = 0.001875 moles

So, 0.001875 moles of NaOH will neutralize 0.001875 moles of CH3COOH. So, excess of CH3COOH left

= 0.003125 moles - 0.001875 moles = 0.00125 moles

Total volume = 25 mL + 15 mL = 40 mL = 0.04 L

So, [CH3COOH] = 0.00125 moles / 0.04 L = 0.03125 M

Now, CH₃COOH $\rightleftharpoons$ H⁺  + CH₃COO^-

Ka = [H⁺] [CH₃COO^-] / [CH₃COOH]

$\Rightarrow$ 1.8 x 10^-5 = (x) (x) / (0.03125 - x)

$\Rightarrow$ 1.8 x 10^-5 = x² / 0.03125 Neglecting x term in denominator, since Ka value is very small

$\Rightarrow$ x² = 5.625 x 10^-7

$\Rightarrow$ x= 7.5 x 10^-4

So, [H⁺] = x= 7.5 x 10^-4

pH = - log  [H⁺] = - log (7.5 x 10^-4) = 3.12

Answer 2

1.

25 mL of 0.125 M CH3COOH

Molarity = Moles / Liter

So, moles of CH3COOH = 0.125 M x 0.025 L = 0.003125 moles

12.5 mL of 0.125 M NaOH

Moles of NaOH = 0.125 M x 0.0125 L = 0.0015625 moles

So, 0.0015625 moles of NaOH will neutralize 0.0015625 moles of CH3COOH. So, excess of CH3COOH left

= 0.003125 moles - 0.0015625 moles = 0.0015625 moles

Total volume = 25 mL + 12.5 mL = 37.5 mL = 0.0375 L

So, [CH3COOH] = 0.0015625 moles / 0.0375 L = 0.0417 M

Now, CH₃COOH $\rightleftharpoons$ H⁺ + CH₃COO^-

Ka = [H⁺] [CH₃COO^-] / [CH₃COOH]

$\Rightarrow$ 1.8 x 10^-5 = (x) (x) / (0.0417 - x)

$\Rightarrow$ 1.8 x 10^-5 = x² / 0.0417 Neglecting x term in denominator, since Ka value is very small

$\Rightarrow$ x² = 7.506 x 10^-7

$\Rightarrow$ x= 8.66 x 10^-4

So, [H⁺] = x= 8.66 x 10^-4  

pH = - log  [H⁺] = - log (8.66 x 10^-4) = 3.06

2.

25 mL of 0.125 M CH3COOH

Molarity = Moles / Liter

So, moles of CH3COOH = 0.125 M x 0.025 L = 0.003125 moles

15 mL of 0.125 M NaOH

Moles of NaOH = 0.15 M x 0.0125 L = 0.001875 moles

So, 0.001875 moles of NaOH will neutralize 0.001875 moles of CH3COOH. So, excess of CH3COOH left

= 0.003125 moles - 0.001875 moles = 0.00125 moles

Total volume = 25 mL + 15 mL = 40 mL = 0.04 L

So, [CH3COOH] = 0.00125 moles / 0.04 L = 0.03125 M

Now, CH₃COOH $\rightleftharpoons$ H⁺  + CH₃COO^-

Ka = [H⁺] [CH₃COO^-] / [CH₃COOH]

$\Rightarrow$ 1.8 x 10^-5 = (x) (x) / (0.03125 - x)

$\Rightarrow$ 1.8 x 10^-5 = x² / 0.03125 Neglecting x term in denominator, since Ka value is very small

$\Rightarrow$ x² = 5.625 x 10^-7

$\Rightarrow$ x= 7.5 x 10^-4

So, [H⁺] = x= 7.5 x 10^-4

pH = - log  [H⁺] = - log (7.5 x 10^-4) = 3.12

Answer 3

1.

25 mL of 0.125 M CH3COOH

Molarity = Moles / Liter

So, moles of CH3COOH = 0.125 M x 0.025 L = 0.003125 moles

12.5 mL of 0.125 M NaOH

Moles of NaOH = 0.125 M x 0.0125 L = 0.0015625 moles

So, 0.0015625 moles of NaOH will neutralize 0.0015625 moles of CH3COOH. So, excess of CH3COOH left

= 0.003125 moles - 0.0015625 moles = 0.0015625 moles

Total volume = 25 mL + 12.5 mL = 37.5 mL = 0.0375 L

So, [CH3COOH] = 0.0015625 moles / 0.0375 L = 0.0417 M

Now, CH₃COOH $\rightleftharpoons$ H⁺ + CH₃COO^-

Ka = [H⁺] [CH₃COO^-] / [CH₃COOH]

$\Rightarrow$ 1.8 x 10^-5 = (x) (x) / (0.0417 - x)

$\Rightarrow$ 1.8 x 10^-5 = x² / 0.0417 Neglecting x term in denominator, since Ka value is very small

$\Rightarrow$ x² = 7.506 x 10^-7

$\Rightarrow$ x= 8.66 x 10^-4

So, [H⁺] = x= 8.66 x 10^-4  

pH = - log  [H⁺] = - log (8.66 x 10^-4) = 3.06

2.

25 mL of 0.125 M CH3COOH

Molarity = Moles / Liter

So, moles of CH3COOH = 0.125 M x 0.025 L = 0.003125 moles

15 mL of 0.125 M NaOH

Moles of NaOH = 0.15 M x 0.0125 L = 0.001875 moles

So, 0.001875 moles of NaOH will neutralize 0.001875 moles of CH3COOH. So, excess of CH3COOH left

= 0.003125 moles - 0.001875 moles = 0.00125 moles

Total volume = 25 mL + 15 mL = 40 mL = 0.04 L

So, [CH3COOH] = 0.00125 moles / 0.04 L = 0.03125 M

Now, CH₃COOH $\rightleftharpoons$ H⁺  + CH₃COO^-

Ka = [H⁺] [CH₃COO^-] / [CH₃COOH]

$\Rightarrow$ 1.8 x 10^-5 = (x) (x) / (0.03125 - x)

$\Rightarrow$ 1.8 x 10^-5 = x² / 0.03125 Neglecting x term in denominator, since Ka value is very small

$\Rightarrow$ x² = 5.625 x 10^-7

$\Rightarrow$ x= 7.5 x 10^-4

So, [H⁺] = x= 7.5 x 10^-4

pH = - log  [H⁺] = - log (7.5 x 10^-4) = 3.12