Question

Calculate the pH of the solution that results from each of the following mixtures.

55.0 mL of 0.16 M HCHO2 with 75.0 mL of 0.13 M NaCHO2

125.0 mL of 0.11 M NH3 with 250.0 mL of 0.11 M NH4Cl

Answer 1

(a): Final volume of the solution after mixing = 55.0 mL + 75.0 mL = 130 mL = 0.130 L

Moles HCHO2 = MxV = 0.16 mol/L x 0.055 L = 0.0088 mol

Hence concentration of HCHO2 after mixing = moles of HCHO2 / Vt = 0.0088mol / 0.130 L = 0.0677 M

Moles NaCHO2 = MxV = 0.13 mol/L x 0.075 L = 0.00975 mol

Hence concentration of NaCHO2 after mixing = moles of NaCHO2 / Vt = 0.00975 mol / 0.130 L = 0.075 M

For HCHO2, pKa = 3.77

HCHO2 and NaCHO2 act as buffer solution whose pH can be calculated from Hendersen equation as

pH = pKa + log[NaCHO2] / [HCHO2]

=> pH = 3.77 + log(0.075 M / 0.0677M) = 3.81 (answer)

(b):

Final volume of the solution after mixing = 125.0 mL + 250.0 mL = 375 mL = 0.375 L

Moles NH3 = MxV = 0.11 mol/L x 0.125 L = 0.01375 mol

Hence concentration of NH3 after mixing = moles of NH3 / Vt = 0.01375mol / 0.375 L = 0.0367 M

Moles NH4Cl = MxV = 0.11 mol/L x 0.250 L = 0.0275 mol

Hence concentration of NH4Cl after mixing = moles of NH4Cl / Vt = 0.0275 mol / 0.375 L = 0.0733 M

For NH3, pKb = 4.75

NH3 and NH4Cl act as buffer solution whose pOH can be calculated from Hendersen equation as

pOH = pKb + log[NH4Cl] / [NH3]

=> pOH = 4.75 + log(0.0733 M / 0.0367M) =