Question

You need to produce a buffer solution that has a ph of 5.42. you already have a solution that contains 10. mmol (millimoles) of acetic acid. how many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? the pka of acetic acid is 4.74.

Answer 1

20 mmol.

Step-by-step explanation

Let
`\text{HA}` denotes acetic acid and
`\text{A}^(-)` acetate ions. Apply the Henderson-Hasselbalch equation:

`\text{pH} = \text{pK}_a - \log{\frac{[\text{HA}]}{[\text{A}^(-)]}}`,

where

• 
  `\text{pH}` the intended pH of the buffer,
• 
  `\text{pK}_a` the pKa constant of the acetic acid, and
• 
  `\frac{[\text{HA}]}{[\text{A}^(-)]}` the ratio between the acetic acid and acetate ion concentrations in the solution.

`\text{pH} = \text{pK}_a - \log{\frac{[\text{HA}]}{[\text{A}^(-)]}}\\\log{\frac{[\text{HA}]}{[\text{A}^(-)]}} = \text{pK}_a - \text{pH}\\\frac{[\text{HA}]}{[\text{A}^(-)]} = e^{\text{pK}_a -\text{pH}}`.

Volume is constant.
`c = n \cdot V`. As a result,

`\frac{n(\text{HA})}{n(\text{A}^(-))} = \frac{V\cdot[\text{HA}]}{V\cdot[\text{A}^(-)]} = \frac{[\text{HA}]}{[\text{A}^(-)]}=e^{\text{pK}_a-\text{pH}}`.

`n(\text{A}^(-)) = \frac{n(\text{HA})}{e^{\text{pK}_a - \text{pH}}} = \frac{10 \;m\text{mol}}{e^(4.74 - 5.42)}=19.7\;m\text{mol}`