Question

A buffer solution consists of 0.00300 M ammonia (NH3) and 0.00500 M ammonium chloride (NH4Cl). What is the change in pH when 0.00100 moles of NaOH are added to one litre of the solution without any change in volume? The pKa of NH4+ is 9.24.

Answer 1

Henderson-Hasselbalch equation for pH of buffer solution :  pH = pKa + log ([A^-]/[HA])

for ammonia buffer solution :

A buffer solution consists of 0.00300 M ammonia (NH3) and 0.00500 M ammonium chloride (NH4Cl).

pH = pKa + log ([NH₃] / [NH₄⁺])

pH = 9.24 + log ([0.00300] / [0.00500]) = 9.02

=====  0.00100 moles of NaOH are added to one litre of the solution without any change in volume :

we will have following reaction :

NH₄⁺ (aq) + OH- (aq) ===> NH₃(aq) + H₂O(l)

new concentrations :

[NH₃] = (( 0.00300 mol/L *1 L ) + 0.00100 mole ))/ 1 L = 0.004 M

[NH₄⁺] = (( 0.00500 mol/L *1 L ) - 0.00100 mole ))/ 1 L = 0.004 M

pH = 9.24 + log ([0.00400] / [0.00400]) = 9.24

Change in pH : pH = + 0.22