If a buffer solution is prepared from 5.15g #NH_4NO_3# and 0.10L of 0.15M #NH_3#, what is the pH of the solution?

2 Answers
Jun 23, 2016

Answer:

#pH=pK_a+log_10{[[A^-]]/[[HA]]}#

Explanation:

It can be shown relatively easily that the #pH# of a weak acid mixed with its conjugate base in appreciable quantities conforms to the so-called #"buffer equation"# as given.

You have not quoted #pK_a# for ammonium ion, but I will take pity on you, because I know that #pK_a#, #NH_4^+ = 9.24#. Note that you are not expected to know these #pK_a# values from memory even as an undergraduate. You ARE expected to know that ammonium ion is a fairly weak acid, and how to manipulate the buffer equation.

Given the #pK_a# it is simply a question of determining the concentrations and plugging in the numbers:

#pH=9.24+log_10{[0.015*mol*L^-1]/[(5.15*g)/(80.04*g/(mol))xx1/(0.10*L)}}# #=# #??#

Note that here I knew that volume of the solution was #0.10*L#. How did I know this?

PS I hope you can see the units in my calculations, because I can't! It is important that the dimensions in the log argument cancel out, which I think they do!

Jun 23, 2016

Answer:

#"pH=8.62"#

Explanation:

The ammonium ion is a weak acid and dissociates:

#NH_(4(aq))^+rightleftharpoonsNH_(3(aq))+H_((aq))^+#

For which:

#K_a=([NH_(3(aq))][H_((aq))^+])/([NH_(4(aq))^+])=5.6xx10^(-10)" ""mol/l"" "# at #25^@"C"#

To find the #"pH"# we need to calculate the concentration of the #H^+# ions.

Rearranging gives:

#[H_((aq))^+]=K_axx([NH_(4(aq))^+])/([NH_(3(aq))])" "color(red)((1))#

These are all equilibrium concentrations. We can set up an #"ICE" # table to find these.

For the initial moles of #NH_3# we use #n=cxxv#:

#:.nNH_(3(aq))=0.15xx0.1=0.015#

The initial moles of #NH_4NO_3# is equal to #m/M_r=5.15/80.04=0.0643#.

From the formula we can see that the initial moles of #NH_(4(aq))^+# must also be #0.0643#.

Now we can set up the #"ICE"# table:

#" "NH_4^+" "rightleftharpoons" "NH_3" "+" "H^+#

#color(red)"I"" "0.0643" "0.015" "0#

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

#color(red)("E")" "(0.0643-x)" "(0.015+x)" "x#

Because the value of #K_a# is so small the value of #x# is much smaller than #0.0643# and #0.015# so we can assume that:

#(0.0643-x)rArr0.0643#

and

#(0.015+x)rArr0.015#

Now we can put the numbers into #color(red)((1))rArr#

#[H_((aq))^+]=5.6xx10^(-10)xx0.0643/0.015=2.4xx10^(-9)" ""mol/l"#

You will note I can use moles and not concentrations as the volume is common to both #[NH_4^+]# and #[NH_3]# so cancels.

This also accounts for any volume change when the solid salt is added to the solution.

#pH=-log[H_((aq))^+]#

#:.pH=-log[2.4xx10^(-9)]=color(red)(8.62)#

In practice you could go straight to #color(red)((1))# but bear in mind the assumptions you have made in approximating equilibrium concentrations to initial ones.

This only works if #K_a# is very small. If not, then you need to use an #"ICE"# table.