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
Michael
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)#

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