Question

1) Calculate the change in pH when 49.0 mL of a 0.780 M solution of NaOH is added to 1.00 L of a solution that is 1.00 M in sodium acetate and 1.00 M in acetic acid.

2) Calculate the pH of 100.0 mL of a buffer that is 0.0650 M NH4Cl and 0.110 M NH3 before and after the addition of 1.00 mL of 6.00 M HNO3.

calculate: pH before and ph After

Answer 1

1. CH₃COOH = CH₃COO^- + H⁺
2. equation

pH = pKa + log[[salt]/[acid]]

pH = 4.76 + log[1M*1L/1M*1L] [pKa of CH₃COOH is 4.76]

= 4.76

By addition of strong base reaction will go forward direction and salt concentration will increase and acid concentration will decrease

Moles of NaOH added = 0.78 M *49 ml

= 0.038 moles

pH = pKa + log[[salt]+[NaOH]/[acid]- [NaOH]]

pH = 4.76 + log[1.038/0.962]

= 4.76 + 0.033

= 4.793

1. NH₃ + H₂O = NH₄⁺ + OH^-

NH₄Cl = NH₄⁺ + Cl^-

[OH] = Kb * [[base]/[salt]] [Kb of NH₃ is 1.8*10^-5]

[OH] = 1.8*10^-5 * [[0.11*0.1]/[0.065*0.1]]

[OH] = 3.04*10^-5

pOH = 4.51

pH = 14 – pOH

= 9.49 (pH before)

By addition of strong acid reaction will go forward direction and salt concentration will increase and base concentration will decrease

Moles of HNO₃ added = 6 M *1 ml

= 0.006 moles

[OH] = Kb * [[salt]+[ HNO₃]/[acid]- [HNO₃]]

= 1.8*10^-5 * [[0.011-0.006]/[0.065+0.006]]

= 1.8*10^-5 *0.4

[OH] = 0.72*10^-5

pOH = 5.14

pH = 8.88 (pH after)