Ammonia (#NH_3#) is a weak base with a #K_b = 1.8 x 10^-5#. What is the balanced chemical equation for the reaction of ammonia with water. Using the I.C.E the method, calculate the pH and ionization of a 1.75 M #NH_3# solution in 2.50 M #NH_4Cl#?

1 Answer
Jan 24, 2017

Answer:

#sf(pH=9.1)#

Explanation:

Ammonia is a weak base and ionises in water:

#sf(NH_3+H_2OrightleftharpoonsNH_4^++OH^-)#

For which:

#sf(K_b=([NH_4^+][OH^-])/([NH_3])=1.8xx10^(-5)color(white)(x)"mol/l")#

These are equilibrium concentrations.

The ICE table in mol/l is:

#" "sf(NH_3" "+" "H_2O" "rightleftharpoons" "NH_4^+" "+" "OH^-)#

#sf(I" "1.75" "2.5" "0)#

#sf(C" "-x" "+x" "+x)#

#sf(E" "(1.75-x") "(2.5+x)" "x#

Because the value of #sf(K_b)# is sufficiently small we can assume that #sf((1.75-x)rArr1.75)# and #sf((2.5+x)rArr2.5)#.

This means we can assume that the initial concentrations are a good enough approximation to those at equilibrium.

There is therefore, no need to set up an ICE table. You can just use the equation for #sf(K_b)# directly and solve for #sf([OH^-])# hence find the pH. You should always state the assumption though.

Rearranging the expression for #sf(K_brArr)#

#sf([OH^-]=K_bxx[[NH_3]]/[[NH_4^+]])#

#:.##sf([OH^-]=1.8xx10^(-5)xx1.75/2.5color(white)(x)"mol/l")#

#sf([OH^-]=1.2xx10^(-5)color(white)(x)"mol/l")#

#sf(pOH=-log[OH^-]=-log(1.2xx10^-5)=4.92)#

#sf(pH+pOH=14)# at 298 K

#:.##sf(pH=14-pOH=14-4.92=9.08)#