Question

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`?

Answer 1

`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)`