The pH of an aqueous solution containing 1 M benzoic acid (pKₐ=4.20) and 1 M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is _________ mL. (given : log 2=0.3)

Solution

Chemistry ma’am · 4 steps

Both solutions are 1 M, so the salt-to-acid ratio in Henderson's equation is just a ratio of volumes.

1. Moles follow volume. Let Vₐ mL of benzoic acid mix with Vₛ mL of sodium benzoate, both 1 M. Each solute ends up in the same total volume, so the ratio of their concentrations equals Vₛ/Vₐ.

   pH = pKₐ + log[salt]/[acid] = pKₐ + log Vₛ/Vₐ

2. Find the ratio. 4.5 = 4.20 + log Vₛ/Vₐ, so log Vₛ/Vₐ = 0.3 = log 2. That gives Vₛ = 2Vₐ.
3. Split 300 mL. Vₐ + Vₛ = 300, so Vₐ + 2Vₐ = 300 and Vₐ = 100 mL. The sodium benzoate takes the other 200 mL.
4. So. The volume of benzoic acid solution in the buffer is 100 mL.

The common slip: 200. Putting acid over salt in the log term flips the ratio and gives 200 mL of benzoic acid, which is really the sodium benzoate volume.

The idea to keep. In Henderson's equation the log term is salt over acid. When pH is above pKₐ, the salt must be in excess, so the acid gets the smaller share.

Answer from the NTA final answer key. Solution by PrepWiser.