Answer to Question #185365 in General Chemistry for Kikikij

Question #185365

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)


1
Expert's answer
2021-04-25T23:10:21-0400

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



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