Question

a) what is the pH of a solution prepared by dissolving 13.1669 g of ammonium sulfate in 850 mL of water?

b) how much 2M ammonium hydroxide is needed t make the pH of 6.25?

please show all work

Answer 1

a) First we write te balanced equation of the ammonium sulfate

(NH₄)₂SO₄ → 2NH₄⁺ + SO₄^2-

We calculate the moles of (NH₄)₂SO₄

n = (13.1669g)/ (132,14 g/mol) = 0,099 mol (NH₄)₂SO₄

if 1 mol of (NH₄)₂SO₄ react with 2 moles of NH₄^+, then 0.099 moles of (NH₄)₂SO₄ form

Xmoles of NH₄⁺= (2 moles of NH₄⁺)x(0.099 moles of (NH₄)₂SO₄)/(1 mol of (NH₄)₂SO₄)

Xmoles of NH₄⁺ = 0.199 moles of NH₄^+,. this moles divided in the Volumen in Liters of solution (0,850 L) will give us a [ ]

M NH₄⁺   = n NH₄^+, / V(L)solution = 0.199 moles of NH₄⁺ / 0.85 L of solution

[NH₄⁺] = 0.23 M

NH₄⁺   ↔ NH₃ + H⁺ Ka = 5.6x10^-10(This is data from textbook)

initial 0.23 0 0

reaction -x +x +x

equilibrium 0.23-x x x

Ka= [NH₃] [H⁺] / [NH₄⁺]

Ka = x²/ (0.23-x) like the Ka is so little we can assume that 0.23-x = 0.23

x²= Ka (0.23) x² = (5.6x10^-10) (0.23) x= 1.1x10^-5 = [H⁺]

pH = -log [H⁺] pH = 4.95

b) Now if we added ammonium hydroxide this will increase the pH for effect of the OH^- ions in the solution.

(NH4)2SO4(aq) + 2NaOH(aq) → Na2SO4(aq) + 2NH3(g) + 2H2O(l)

Now we do the opposite at what we calculate before.

First we calculate the [H⁺] from the pH

[H⁺] = 10^-pH   [H⁺] = 5.6x10^-7 = [NH₃] (from the equation of the dissociation of the NH₄⁺

now we do the stequiometric relation and find the mass af NaOH