An aqueous solution contains 0.466 M ammonia (NH3).

Question

An aqueous solution contains 0.466 M ammonia (NH₃).  

How many mL of 0.278 M hydrochloric acid would have to be added to 225 mL of this solution in order to prepare a buffer with a pH of 9.180. 

Answer 1

pOH = pKb + log [salt]/[base]

pKb NH3 = 4.74

pH = 9.18

pOH = 14 - 9.18 = 4.82

4.82 = 4.74 + log [salt]/[base]

log [salt]/[base] = 0.08

[salt]/[base] = 1.2/1

NH3 + HCl ===> NH4Cl

0.105.....x...............0.........Initial

-x..........-x..............+x........Change

0.105-x...0..............x.........Equilibrium

x/0.105-x = 1.2

0.126 - 1.2x = x

2.2x = 0.126

x = 0.0573

0.0573 mol HCl = (x L)(0.278 mol/L)

x = 0.206 L = 206 ml of HCl needed