Question

Calculate the pH of a buffer when 0.010 moles of NaOH is added to 100. mL solution that is 0.20 M sodium pentanoate (C4H9COONa) and 0.20 M pentanoic acid (C4H9COOH). pKa(C4H9COOH)

Answer 1

5.30

Step-by-step explanation:

The pH of a buffer given by the Henderson-Hasselbach equation can be expressed as:

`pH = pKa + log ([salt])/([Acid])`

Given that:

Volume of the solution = 100.0 mL

To liters; the volume of the solution be 0.1 L

The concentration of
`C_4H_9COONa` = 0.20 M

The concentration of
`C_4H_9COONa` = molarity × volume

The concentration of
`C_4H_9COONa` = 0.20 mol/L × 0.1 L

The concentration of
`C_4H_9COONa` = 0.02 mol

The concentration of
`C_4H_9COOH` = 0.20 M

The concentration of
`C_4H_9COOH` = 0.20 mol/L × 0.1 L

The concentration of
`C_4H_9COOH` = 0.02 mol

However, the number of moles of NaOH added = 0.01 moles

Now; The ICE table can be computed as:

C₄H₉COOH + OH⁻ ⇄ C₄H₉COO⁻ + H₂O

Initial 0.02 0.01 0.02

Change - 0.01 -0.01 +0.01 +0.01

Equilibrium 0.01 - 0.03 0.01

Recall that the pH of
`C_4H_9COOH` = 4.82

∴

`pH = 4.82 + log \begin {pmatrix} ((0.03)/(0.1) )/( (0.01)/(0.1) ) \end {pmatrix}`

`pH = 4.82 + log \begin {pmatrix} (0.3 )/(0.1 ) \end {pmatrix}`

pH = 4.82 + log ( 3 )

pH = 4.82 + 0.4771

pH = 5.2971

pH ≅ 5.30