Question

Calculate the pH of a 1 L solution made by mixing 0.25mol of NH3 and 0.35mol NH4Cl. (Kb = 1.8x10^- 5). What is the new pH if 0.15 mol HCl is added to the solution?

Answer 1

• Calculate the pH of a 1 L solution made by mixing 0.25mol of NH3 and 0.35mol NH4Cl

V_sol=1 L

n_NH3=0.25mol ; n_NH4⁺=0.35mol

   Kw=1x10^-14 (water dissociation constant) --------->

Henderson Hasselbach equation pH = pKa + log(base/acid)

pH = pKa + log([NH₃ ][NH₄⁺]) ; where pKa= -LogKa

[NH₃ ]= n_NH3/V_sol =0.25mol/ 1L= 0.25 mol/L ;   [NH₄⁺]= n_NH4⁺/V_sol=0.35mol/1L= 0.35 mol/L

pH= -log(5.56x10^-10) + Log(0.25/0.35)

pH= 9.11

• What is the new pH if 0.15 mol HCl is added to the solution?

NH₄⁺ <-------------->H⁺ + NH₃

Initial 0.35mol 0 0.25mol

Adding Acid 0.15mol -0.15mol

After Acid 0.50mol 0.1mol

pH = pKa + log([NH₃ ][NH₄⁺])

pH= -Log(5.56x10^-10) + Log(0.1/0.50)

pH= 8.56