Answer to Question #160497 in Inorganic Chemistry for amirul

Question

solution z is prepared by mixing 0.15dm3 of solution X and 0.35 dm3 of solution Y. Solution X contains 0.12 mole of ammonia, NH3 while solution Y contains 0.09 moles of ammonium chloride, NH4CL. Determine the pH of solution Z after the addition of 0.01 mole of hydrochloric acid, HCl.( Given the pKb of NH3 is 4.75)

Answer 1

NH₃ + NH₄Cl - buffer solution

NH₃ + HCl = NH₄Cl

0.12 mole 0.01 mole 0.01 mole

HCl - limiting reagent

n(NH₃) = 0.12 mole - 0.01 mole = 0.11 mole

n(NH₄Cl) = 0.09 mole + 0.01 mole = 0.10 mole

V(solution) = 0.15 L + 0.35 L = 0.5 L

C(NH₃) = n(NH₃) / V(solution) = 0.11 mole / 0.5 L = 0.22 M

C(NH₄Cl) = n(NH₄Cl) / V(solution) = 0.10 mole / 0.5 L = 0.20 M

pOH = pK_b + log (molarity of salt / malarity of base) = 4.75 + log (0.20 / 0.22) = 4.71

pH = 14 - 4.71 = 9.29