Question

What volumes of 50 mM `CH_3COOH` and 30 mM `NaCH_3COO` solutions are needed to prepare 2L of 20 mM acetate buffer pH 3.6? (`pK_a` for acetic acid is 4.74)

Answer 1

WARNING! Long answer! You need to mix 750 mL of acetic acid with 90 mL of sodium acetate and add enough water to make 2 L of buffer.

Explanation:

The strategy to follow is

1) Write the chemical equation for the buffer.
2) Calculate the concentration ratio of `"CH"_3"COOH"` and `"CH"_3"COONa"`.
3) Calculate the molar ratio of `"CH"_3"COOH"` and `"CH"_3"COONa"` .
4) Calculate the moles of `"CH"_3"COOH"` and `"CH"_3"COONa"`.
4) Calculate the volumes of `"CH"_3"COOH"` and `"CH"_3"COONa"`.

1) Chemical equation

`"HA" + "H"_2"O" ⇌ "H"_3"O"^(+) + "A"^"-"`; `"p"K_text(a) = 4.74`

2) Calculate the concentration ratio

We use the Henderson-Hasselbalch equation to calculate the concentration ratio.

`"pH" = "p"K_"a" + log(("[A"^"-""]")/"[HA]")`

`3.6 = 4.74 + log(("[A"^"-""]")/"[HA]")`

`log(("[A"^"-""]")/"[HA]") = "3.6 - 4.74" = -1.14`

`log(("[HA]")/("[A"^"-""]")) = 1.14`

`("[HA]")/("[A"^"-""]") = 10^1.14 = 13.8`

`"[HA]" = "13.8[A"^"-""]"`

3) Calculate the molar ratio

Since both the acetic acid and the sodium acetate are in the same solution, the ratio of their concentrations is the same as the ratio of their moles. That is,

`n_"HA" = 13.8n_"A⁻"`

4) Calculate the moles

`"Total moles" = 2 "L" × "20 mmol"/(1 "L") = "40 mmol"`

`n_"HA" + n_"A⁻" = 13.8n_"A⁻" + n_"A⁻" = 14.8n_"A⁻" = "40 mmol"`

`n_"A⁻" = "40 mmol"/14.8 = "2.70 mmol"`

∴ `n_"HA" = ("40-2.70) mmol" = "37.3 mmol"`

5) Calculate the volumes

`V_"HA" = 37.3 color(red)(cancel(color(black)("mmol HA"))) × "1 L HA"/(50 color(red)(cancel(color(black)("mmol HA")))) = "0.75 L HA" = "750 mL HA"`

`V_"HA" = 2.70 color(red)(cancel(color(black)("mmol A⁻"))) × "1 L A⁻"/(30 color(red)(cancel(color(black)("mmol A⁻")))) = "0.090 L A⁻" = "90 mL A⁻"`

Thus, you mix 750 mL of the acetic acid with 90 mL of the sodium acetate and add enough water to make 2 L of pH 3.6 buffer.