Question

What is the pH change when 22.4 mL of .07 M NaOH is added to 63.6 mL of a buffer solution consisting of .124 M NH3 and 0.153 M NH4Cl? (Ka for ammonium ion is 5.6x10^-10.)

Answer 1

Lets calculate the initial pH

Ka = 5.6*10^-10

pKa = - log (Ka)

= - log(5.6*10^-10)

= 9.252

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.252+ log {0.124/0.153}

= 9.161

Lets calculate the final pH

mol of NaOH added = 0.07M *22.4 mL = 1.568 mmol

NH4+ will react with OH- to form NH3

Before Reaction:

mol of NH3 = 0.124 M *63.6 mL

mol of NH3 = 7.8864 mmol

mol of NH4+ = 0.153 M *63.6 mL

mol of NH4+ = 9.7308 mmol

after reaction,

mol of NH3 = mol present initially + mol added

mol of NH3 = (7.8864 + 1.568) mmol

mol of NH3 = 9.4544 mmol

mol of NH4+ = mol present initially - mol added

mol of NH4+ = (9.7308 - 1.568) mmol

mol of NH4+ = 8.1628 mmol

Ka = 5.6*10^-10

pKa = - log (Ka)

= - log(5.6*10^-10)

= 9.252

since volume is both in numerator and denominator, we can use mol instead of concentration

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.252+ log {9.454/8.163}

= 9.316

pH change = |final pH - initial pH|

= |9.316 - 9.161|

= 0.155

Answer: 0.155