Question

1. Calculate the pH of a solution that is made by mixing 0.80 L of 0.30 M formic acid (HCOOH) and 0.90 L of 0.20 M sodium formate (NaHCOO). Assume volumes are additive. (Ka, HCOOH = 1.77 * 10^-4 )

2. A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80M sodium hydroxide. What was the pH of the solution after 50.0 mL of the NaOH solution were added? The Kb of ammonia is 1.76 * 10^-5 .

Answer 1

1)

Ka = 1.77 x 10^-4

pKa = 3.75

moles of formic acid = 0.8 x 0.30 = 0.24

moles of sodium formate = 0.90 x 0.20 = 0.18

pH = pKa + log [salt / acid]

    = 3.75 + log [0.18 / 0.24]

pH = 3.62

2)

pKb = 4.75

pKa = 9.25

moles of NH4+ = 200 x 0.40 = 80

moles of NaOH = 50 x 0.8 = 40

NH4+   +   OH-    ------------> NH3 + H2O

80              40                          0             0

40              0                            40

this is half - equivalence point.

here pH = pKa

pH = 9.25