Question

Solve an equilibrium problem (using an ICE table) to calculate the pH of of each solution: a solution that is 0.15 M in HCHO2 and 0.15 M in NaCHO2

and a solution that is 0.14 M in NH3 and 0.22 M in NH4Cl

Answer 1

Concentration of HCOOH = 0.15M Concentration of HCOONa = 0.15M Ka of HCOOH is 1.77 times 10^-4 pKa = -log (1.77 times 10^-4) = 3.75 According to Henderson-Hasselbalch equation: pH = pKa + log [HCOONa]/[HCOOH] pH = 3.75 + log (0.15)/(0.15) pH = 3.75 + log(1) = 3.75 - 0.00 = 3.75 Hence pH of solution is 3.69 Concentration of NH_3 = 0.14M Concentration of NH_4 Cl = 0.22 M Kb of NH_3 is 1.78 times 10^-5 pKb = -log (1.78 times 10^-5) = 4.75 According to Henderson-Hasselbalch equation: pOH = pKb + log [NH_4 Cl]/[NH_3] pOH = 4.75 + log (0.22)/(0.14) pOH = 4.75 + log (1.5) = 4.75 + 0.196 = 4.95 pH = 14 - 4.95 = 9.05 Hence pH of solution is 9.05