Question

Suppose

0.891g

of ammonium acetate is dissolved in

50.mL

of a

0.10M

aqueous solution of sodium chromate.

Calculate the final molarity of ammonium cation in the solution. You can assume the volume of the solution doesn't change when the ammonium acetate is dissolved in it.

Answer 1

Molar mass of ammonium acetate = 77.0825 g mol^-1

Moles of amm. acetate = 0.891/77.0825 =0.01155

molarity of amm. acetate = 0.01155x1000/50 = 0.2311 mol L^-1

However, ammonium acetate forms a buffer solution.

It undergoes hydrolysis to form ammonium hydroxide and aceticacid both are weak electrolytes

pH = 1/2(pK_w +pK_a - pKb)

Here K_a= K_b  = 1.8 x10^-5

NH_{4⁺ +} OH^- ------------- NH4OH

0.2311-x x x

1/Kb = x/x(0.2311-x) = 1/0.2311-x

Since the value of Kb is less compared to x

Molarity of Ammonium ions = 0.2311 mol L^-1