Question #dbab5

1 Answer
Apr 25, 2017

Answer:

#sf(pH=9.9)#

Explanation:

The ammonium ion ionises:

#sf(NH_4^+rightleftharpoonsNH_3+H^+)#

For which:

#sf(K_a=([NH_3][H^+])/([NH_4^+])=5.6xx10^(-10)color(white)(x)"mol/l")#

When #sf(OH^-)# ions are added the following takes place:

#sf(NH_4^++OH^(-)rarrNH_3+H_2O)#

The no. of moles added is given by:

#sf(n_(OH^-)=cxxv=0.75xx20.0/1000=0.015)#

This means, from the equation, that the number of moles of #sf(NH_4^+)# consumed = 0.015

#:.# The number of moles remaining is given by:

#sf(n_(NH_4^+)=0.025-0.015=0.01)#

From the equation you can see that the number of moles of #sf(NH_3)# formed = 0.015.

The total moles of #sf(NH_3)# is given by:

#sf(n_(NH_3)=0.04+0.015=0.055)#

The small value of #sf(K_a)# means we can assume that these are the moles present at equilibrium.

Rearranging the expression for #sf(K_a)# we get:

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

Since the total volume is common to both salt and base we can use moles directly:

#sf([H^+]=5.6xx10^(-10)xx0.01/0.055color(white)(x)"mol/l")#

#sf([H^+]=1.018xx10^(-10)color(white)(x)"mol/l")#

#sf(pH=-log[H^+]=-log[1.018xx10^(-10)])#

#sf(pH=9.99)#