Question

What is the net ionic equation for the reaction between ammonia and ammonium ions in a buffer solution? What is the pH of a buffer prepared by mixing 150 mL of 0.60 mol/L ammonium chloride and 100 mL of 0.90 mol/L aqueous ammonia (`"p"K_text(b) = 4.75`)?

Answer 1

There is no net ionic reaction between `"NH"_4^+` and `"NH"_3`.

Explanation:

This `color(red)("is")` your buffer solution.

In this problem, you already have the buffer components — the weak base `"NH"_3` and its conjugate acid `"NH"_4^+`.

The only net ionic reaction that you need is the equation for the ionization of ammonia:

`"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"`; `"p"K_text(b) = 4.75`

You still have to calculate the moles of each component.

`"Moles of NH"_4^"" = 0.150 cancel("L") × "0.60 mol"/(1 cancel("L")) = "0.090 mol"`

`"Moles of NH"_3 = 0.100 cancel("L") × "0.90 mol"/(1 cancel("L")) = "0.090 mol"`

The Henderson-Hasselbalch Equation is;

`"pOH" = "p"K_"b" + log(("[NH"_4^(+)"]")/("[NH"_3"]"))`

`"pOH" = 4.75 + log((cancel("0.090 mol"))/(cancel("0.090mol"))) = 4.75 – 0.00 = 4.75`

`"pH = 14.00 – pOH = 14.00 – 4.75 = 9.25"`