In: Chemistry
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.)
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