Question

Calculate how many moles of base it takes to change the pH of the following buffered solution by 0.10 pH unit.

0.100 L of 0.25 M NH3 + 0.050 L of 0.25 M HCl

（）mol

Answer 1

NH3 moles = M x V = 0.25 x 0.1 = 0.025

Moles of HCl = 0.25 x 0.05 = 0.0125 = moles of H+

we have reaction NH3 (aq) + H+ (aq) <---> NH4+ (aq)

after reatcing with H+ , we have NH3 moles = 0.025-0.0125 = 0.0125

NH4+ moles formed = H+ moles reacted = 0.0125

pH = pkb+ log ( Nh4+) moles / ( NH3 moles)

= 4.75 + log ( 0.0125/0.0125)

= 4.75

pH change = 0.1 hence pOH change is also 0.1, hence pOH = 4.75 -0.1 = 4.74

hence now 4.74 = 4.75 + log ( NH4+ / NH3 moles)

( Nh4+ moles/NH3moles) = 0.7943

NH4+ moles = 0.7943 ( NH3 moles) ...............(1)

Let OH- moles added = m

then NH3 moles = 0.0125+m ,   NH4+ moles = 0.0125-m     ( since OH- + NH4+ ---> NH3 + H2O )

by eq (1) we get

(0.0125-m) = 0.7943 ( 0.0125+m)

m = 0.0014 = moles of base added   to get pH change of 0.1