Question

Calculate the pH of the following aqueous solutions?

Calculate the pH of the following aqueous solutions?
a) 0.1 M ammonium chloride.

b) 100 ml of 0.1 M solution of ammonium chloride to which 100 ml of
0.1 M solution of sodium hydroxide.

Assume that all volumes are additive. Kb (ammonia) = 1.8 x 10-5

Answer 1

Warning! Long Answer. a) pH = 5.13; b) pH = 11.0

Explanation:

For a):

Ammonium chloride, `NH_4Cl` dissolves in solution to form ammonium ions `NH_4^(+)` which act as a weak acid by protonating water to form ammonia, `NH_3(aq)` and hydronium ions `H_3O^(+)(aq)`:

`NH_4^(+)(aq)+H_2O(l) -> NH_3(aq)+H_3O^(+)(aq)`

As we know the `K_b` for ammonia, we can find the `K_a` for the ammonium ion. For a given acid/base pair:

`K_a times K_b=1.0 times 10^-14` assuming standard conditions.

So, `K_a(NH_4^(+))=(1.0 times 10^-14)/(1.8 times 10^-5)=5.56 times 10^-10`

Plug in the concentration and the `K_a` value into the expression:

`K_a=([H_3O^(+)] times (NH_3])/([NH_4^(+)])`

`5.56 times 10^-10~~([H_3O^(+)] times [NH_3])/([0.1])`

`5.56 times 10^-11=[H_3O^(+)]^2`

(as we can assume that one molecule hydronium must form for every one of ammonia that forms. Also, `K_a` is small, so `x ≪ 0.1`.)

`[H_3O^(+)]=7.45 times 10^-6`

`pH=-log[H_3O^(+)]`

`pH=-log(7.45 times 10^-6)`

`pH approx 5.13`

For b):

(i) Determine the species present after mixing.

The equation for the reaction is

`color(white)(mmmmm)"OH"^"-" + "NH"_4^"+" -> "NH"_3 + "H"_2"O"`
`"I/mol": color(white)(mll)0.010 color(white)(mll)0.010color(white)(mol/L)0`
`"C/mol": color(white)(m)"-0.010"color(white)(ml)"-0.010"color(white)(m)"+0.010"`
`"E/mol": color(white)(mll)0color(white)(mmmm)0color(white)(mmml)0.010`

`"Moles of OH"^"-" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol"`

`"Moles of NH"_4^"+" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol"`

So, we will have 200 mL of an aqueous solution containing 0.010 mol of ammonia, and the pH should be higher than 7.

(ii) Calculate the pH of the solution

`["NH"_3] = "0.010 mol"/"0.200 L" = "0.050 mol/L"`

The chemical equation for the equilibrium is

`"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"`

Let's re-write this as

`"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"`

We can use an ICE table to do the calculation.

`color(white)(mmmmmmmll)"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"`
`"I/mol·L"^"-1":color(white)(mll)0.050color(white)(mmmmmll)0color(white)(mmm)0`
`"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmll)"+"xcolor(white)(mll)"+"x`
`"E/mol·L"^"-1":color(white)(m)"0.050-"xcolor(white)(mmmmm)xcolor(white)(mmm)x`

`K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = x^2/("0.050-"x) = 1.8 × 10^"-5"`

Check for negligibility:

`0.050/(1.8 × 10^"-5") = 3 × 10^3 > 400`. ∴ `x ≪ 0.050`

`x^2/0.050 = 1.8 × 10^"-5"`

`x^2 = 0.050 × 1.8 × 10^"-5" = 9.0 × 10^"-7"`

`x = 9.5 × 10^"-4"`

`["OH"^"-"] = 9.5 × 10^"-4" color(white)(l)"mol/L"`

`"pOH" = -log(9.5 × 10^"-4") = 3.0`

`"pH = 14.00 - pOH = 14.00 - 3.0" = 11.0`