Answer to Question #185365 in General Chemistry for Kikikij

Question

How many moles of NH,CI must be added to 1.5L of 0.2M solution of NH, to form a buffer whose P

is 9.00 ?(Kb = 1.8 × 105)

Answer 1

pOH=pKb+log([conjugate acid]/[weak base])

To do that, you need to know the value of the base dissociation constant, Kb

, for ammonia

Kb=1.8 ×10^-5

So, use the given pH to find the pOH

 of the solution

pOH=14−pH

=14−9.00=5.00

This means that you have

5.00=−log(1.8 ×10^-5)+log([NH⁺₄][NH₃])

5.00=4.74+log([NH⁺₄][NH₃])

This is equivalent to

log([NH⁺₄][NH₃])

=0.26

To get rid of the log, use

[NH⁺₄][NH₃]

=10

0.26

This wil get you[NH⁺₄][NH₃]=1.8197

This tells you tha the ratio between the concentration of conjugate acid and the concentration of weak base must be equal to 1.8197

.

The concentration of ammonium ions will thus be

[NH⁺₄]=1.8197[NH₃][NH⁺₄]

=1.8197×0.930 M

=1.6923 M

Now, ammonium chloride dissociates completely in aqueous solution to give

NH₄Cl(aq]→NH⁺

4(aq] + Cl^-

(aq]

Notice that 1

 mole of ammonium chloride produces 1

 mole of ammonium ions in solution. This means that the molarity of the ammonium chloride will be equal to that of the ammonium ions

[NH₄Cl]=[NH⁺₄]=1.6923 M