A solution is prepared by adding 0.040 moles of HCl to 100 mL of a 0.50 M ammonia solution. Ka(NH4+) = 5.6 x 10-10. Assuming no volume change upon addition of the HCl, what is the resulting pH?

1 Answer
Oct 19, 2017

pH = 8.65

Explanation:

The following neutralisation takes place:

sf(NH_3+HClrarrNH_4^+Cl^-)

The initial no. of moles of ammonia is given by:

sf(n=cxxv=0.50xx100/1000=0.05)

The no. of moles of HCl added = sf( 0.04)

This means the no. of moles of sf(NH_4^+) formed sf(= 0.04)

The no. of moles of ammonia remaining sf(=0.05-0.04=0.01)

The ammonium ions undergo hydrolysis:

sf(NH_4^+rightleftharpoonsNH_3+H^+)

sf(K_a=([NH_3][H^+])/([NH_4^+])=5xx10^(-10))

These are equilibrium concentrations. Because of the small value of sf(K_a) we can assume that they approximate to the initial values.

Rearranging:

sf([H^+]=K_axx([NH_4^+])/([NH_3]))

We can use moles instead of concentrations since the total volume is common. We don't even need to assume zero volume change on adding HCl as the question states.

:.sf([H^+]=5.6xx10^(-10)xx0.04/0.01=2.24xx10^(-9)color(white)(x)"mol/l")

sf(pH=-log[H^+]=-log(2.24xx10^(-9)) =8.65)