Question

Find the mass of sodium formate that must be dissolved in 380.0 cm3

of a 0.90 M
solution of formic acid to prepare a buffer solution with pH
= 3.70.

Answer 1

To prepare a buffer solution with a pH of 3.70 using a 0.90 M solution of formic acid, 20.70 g of sodium formate must be dissolved in 380.0 cm³.

To find the mass of sodium formate needed to prepare a buffer solution with a specific pH, we'll use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of its acid and conjugate base. The equation is given by:

`\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{A⁻}]}{[\text{HA}]}\right) \]`

For a solution of formic acid
`(\(HA\))` and sodium formate
`(\(A^-\))` the pKa for formic acid is 3.75.

Given pH = 3.70 and
`\(\text{pKa} = 3.75\)`, we can rearrange the Henderson-Hasselbalch equation to solve for
`\([\text{A⁻}]/[\text{HA}]\)`:

`\[ \frac{[\text{A⁻}]}{[\text{HA}]} = 10^{\text{pH} - \text{pKa}} \]`

`\[ \frac{[\text{A⁻}]}{[\text{HA}]} = 10^(3.70 - 3.75) \]`

`\[ \frac{[\text{A⁻}]}{[\text{HA}]} = 0.56 \]`

Now, we can use the molarity
`(\(M\))` and volume
`(\(V\))` relationship to find the moles of formic acid
`(\(HA\))` in 380.0 cm³:

`\[ \text{moles of } HA = M * V \]`

`\[ \text{moles of } HA = 0.90 \, \text{mol/L} * (380.0 * 10^(-3) \, \text{L}) \]`

`\[ \text{moles of } HA = 0.342 \, \text{mol} \]`

Since
`\(\frac{[\text{A⁻}]}{[\text{HA}]} = 0.56\), the moles of \(A^-\) is \(0.56 * 0.342 \, \text{mol} = 0.191 \, \text{mol}\).`

Now, calculate the mass of sodium formate
`(\(A^-\))` using its molar mass:

`\[ \text{Mass of } A^- = \text{moles of } A^- * \text{Molar mass of } A^- \]`

`\[ \text{Mass of } A^- = 0.191 \, \text{mol} * 68.01 \, \text{g/mol} \]`

`\[ \text{Mass of } A^- \approx 12.97 \, \text{g} \]`

Therefore, approximately 12.97 grams of sodium formate must be dissolved in 380.0 cm³ of a 0.90 M solution of formic acid to prepare a buffer solution with pH 3.70.