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# Calculating the effect of a common ion on concentrations, pH and percent dissociation

In 0.15 M NH3, the pH is 11.21 and the percent dissociation is 1.1%. Calculate the concentrations of all species present, the pH, and the percent dissociation of ammonia in a solution that is 0.15 M in NH3 and 0.45 in NH4Cl.
Kb=1.8 x 10-5

Solution:

The principal reaction is proton transfer to NH3 from H2O :
${ NH }_{ 3(aq) }+{ H }_{ 2 }{ O }_{ (l) }\rightarrow { NH }_{ 4(aq) }^{ + }+{ OH }_{ (aq) }^{ + }$

Since NH4+ ions come both from the NH4Cl present initially (0.45M) and from the reaction of NH3 with H2O, the concentrations of the species involved in the principal reaction are as follows:

The equilibrium equation for the principal reaction is
${ K }_{ b }=\frac { { \left[ { NH }_{ 4 }^{ + } \right] }\left[ { OH }^{ - } \right] }{ \left[ { NH }_{ 3 } \right] } \\ \\ 1.8\times { 10 }^{ -5 }=\frac { \left( 0.45+x \right) \left( x \right) }{ \left( 0.15-x \right) } \approx \frac { 0.45x }{ 0.15-x }$

We assume x is negligible compared to 0.45 and 0.15 because:
– The equilibrium constant Kb is small
– The equilibrium is shifted to the left by the common – ion effect

Therefore, $x=\frac { 1.8\times { 10 }^{ -5 }\times 0.15 }{ 0.45 } =6\times { 10 }^{ -6 }$

$\left[ { OH }^{ - } \right] =6\times { 10 }^{ -6 }M\\ \\ \left[ { NH }_{ 3 } \right] =0.15-6\times { 10 }^{ -6 }M=0.15M$
$\left[ { NH }_{ 4 }^{ + } \right] =0.45-6\times { 10 }^{ -6 }=0.45M$

Thus, the assumption concerning the size x is justified
The H3O+ concentration and the pH are

$\left[ { H }_{ 3 }{ O }^{ + } \right] =\frac { { K }_{ w } }{ \left[ { OH }^{ - } \right] } =\frac { { 1\times 10 }^{ -14 } }{ 6\times { 10 }^{ -6 } } =1.7\times { 10 }^{ -9 }M$ $pH=-log(1.7\times { 10 }^{ -9 })=8.77$

Percentage of dissociation of ammonia is
$=\frac { { \left[ { NH }_{ 3 } \right] }_{ dissociated } }{ { \left[ { NH }_{ 3 } \right] }_{ initial } }$x100%
$=\frac { 6\times { 10 }^{ -6 } }{ 0.15 }$x100%=0.004%