Consider the titration of 300.0 mL of #0.300 MNH_3 (K_b = 1.8 xx 10^-5)# with #0.300 M HNO_3#. What is the pH at the equivalence point?

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Answer 1 · 2016-07-11T21:53:24

#sf(pH=5.15)#

At the equivalence point the reaction is complete:

#sf(NH_(3(aq))+HNO_(3(aq))rarrNH_4NO_(3(aq))#

The number of moles of #sf(NH_3)# reacting = #sf(0.3xx300/1000=0.09)#

So the number of moles of #sf(NH_4^+)# formed = #sf(0.09)#

Now we can set up an ICE table:

#" "##sf(NH_4^(+)" "rightleftharpoons" "NH_3" "+" "H^+)#

#sf(color(red)(I) " " 0.09" "0" "0 )#

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

#sf(color(red)(E)" "(0.09-x)" "x" "x)#

#sf(K_a=([NH_3][H^+])/([NH_3])#

We can find #sf(K_a)# using:

#sf(K_w=K_axxK_b)#

#:.##sf(K_a=(10^(-14))/(1.8xx10^(-5))=5.55xx10^(-10)" ""mol/l")# at #sf(25^@C)#

The concentrations are at equilibrium so we can write:

#sf(K_a=(x.x)/((0.09-x))=5.55xx10^(-10))#

Because the value of #sf(K_a)# is so small we can assume that #sf((0.09-x)rArr0.9)#.

#:.##sf((x^2)/0.09=5.55xx10^(-10))#

#:.##sf(x^2=4.99xx10^(-11))#

#sf(x=7.07xx10^(-6)=[H^+]_(eqm))#

#sf(pH=-log[H^+]_(eqm)=-log[7.07xx10^(-6)])#

#sf(pH=5.15)#

This is an example of "salt hydrolysis". When a strong acid and a weak base produce a salt the final product is slightly acidic.