Question

Suppose you want to prepare a buffer with a pH of 4.59 using formic acid. What ratio of [sodium formate]/[formic acid) do you need to make this buffer? Formic acid has a K, of 1.8x10 4 Enter your answer to three significant figures.

Answer 1

Answer:

7.08

Explanation:

To solve this problem we'll use the Henderson-Hasselbach equation:

• pH = pka + log$\frac{[A^-]}{[HA]}$

Where $\frac{[A^-]}{[HA]}$ is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:

• pka = -log (1.8x10⁻⁴) = 3.74

We input the data:

• 4.59 = 3.74 + log$\frac{[A^-]}{[HA]}$

And solve for $\frac{[A^-]}{[HA]}$:

• 0.85 = log$\frac{[A^-]}{[HA]}$

• $10^{(0.85)}$=$\frac{[A^-]}{[HA]}$

• $\frac{[A^-]}{[HA]}$ = 7.08

Answer 2

The ratio of [sodium formate] to [formic acid] needed to make the buffer is 7.08

How to determine the pKa

• Equilibrium constant (Ka) = 1.8×10¯⁴
• pKa =?

pKa = –Log Ka

pKa = –Log 1.8×10¯⁴

pKa = 3.74

How to determine the ratio of [sodium formate] to [formic acid]

• pH = 4.59
• pKa = 3.74
• Ratio of [sodium formate] to [formic acid] =?

pH = pKa + Log[sodium formate]/[formic acid]

4.59 = 3.74 + Log[sodium formate]/[formic acid]

Collect like terms

4.59 – 3.74 = Log[sodium formate]/[formic acid]

0.85 = Log[sodium formate]/[formic acid]

Take the anti-log of 0.85

[sodium formate]/[formic acid] = anti-log 0.85

[sodium formate]/[formic acid] = 7.08

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