Question

6. (a) Calculate the pH of the 0.39 M NH3/ 0.73 M NH4Cl buffer system. pH = (b) What is the pH after the addition of 20.0 mL of 0.075 M NaOH to 80.0 mL of the buffer solution? pH =

Answer 1

a)

Henderson- Hasselbalch equation is

pH = pKa + log([A^-]/[HA])

where,

A^- = conjucate base , NH3

HA = weak acid , NH4⁺

pKa of NH4⁺ = 9.25

pH = 9.25 + log(0.39M/0.73M)

pH = 9.25 -0.27

pH = 8.98

b)

Initial moles of NH3 = (0.39mol/ 1000ml) × 80ml = 0.0312mol

Initial moles of NH4⁺ = ( 0.73mol/1000ml) × 80ml = 0.0584mol

moles of NaOH added = ( 0.075mol/1000ml) × 20.0ml = 0.0015mol

NaOH react with weak acid NH4⁺

NH4⁺ + OH^- --------> NH3 + H2O

0.0015moles of OH^- react with 0.0015moles of NH4⁺ to give 0.0015moles of NH3

after addition of NaOH

moles of NH3 = 0.0312mol + 0.0015 = 0.0327mol

moles of NH4⁺ = 0.0584mol - 0.0015 = 0.0569mol

Total volume = 80.0ml + 20.0ml = 100ml

[NH3] = ( 0.0327mol/100ml) × 1000ml = 0.327M

[NH4⁺] = (0.0569mol/100ml) × 1000ml = 0.569M

Applying Henderson- Hasselbalch equation

pH = pKa + log([NH3]/[NH4⁺])

pH = 9.25 + log( 0.327M/0.569M)

pH = 9.25 - 0.24

pH = 9.01