You need to prepare 100.0mL of a pH=4.00 buffer solution using 0.100 M benzoic acid (pKa=4.20) and 0.240 M sodium benzoate. How much of each solution should be mixed to prepare the buffer?

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Apr 10, 2017

#"79.18 mL aqueous benzoic acid"#
#"20.82 mL aqueous sodium benzoate"#


Well, we know the final #"pH"#, so the first thing we can solve for is the weak-base/weak-acid ratio via the Henderson-Hasselbalch equation (we are in the buffer region, so this equation works!).

#"pH" = "pKa" + log\frac(["A"^(-)])(["HA"])#

where #"A"^(-)# is the benzoate and #"HA"# is the benzoic acid.

The ratio is then...

#4.00 = 4.20 + log\frac(["A"^(-)])(["HA"])#

#=> \frac(["A"^(-)])(["HA"]) = 10^(4.00 - 4.20) = 0.6310#

It makes sense; #"pH"# #<# #"pKa"#, so the solution is more acidic than if there were equal quantities of benzoate and benzoic acid. Hence, there is more weak acid than weak base.

This ratio, however, is NOT the starting concentrations given to you. It's the ratio in the buffer, i.e. the ratio after the buffer has been finalized.

There is a dilution!

Since the solution has only one total volume, the total volumes cancel out for the dilution, and we only need to determine the initial volumes necessary to accomplish the #ul("mol":"mol")# ratio of #0.6310#.

#=> 0.6310 = ("0.240 M" xx V_(A^(-))/(cancel(V_(t ot))))/("0.100 M" xx (V_(HA))/(cancel(V_(t ot))))#

#= ("0.240 M" xx V_(A^(-)))/("0.100 M" xx V_(HA))#

Now, we do actually have to assume something. We assume that the volumes are additive, so that we can find, say, #V_(A^(-))# in terms of #V_(HA)#. We know that the total volume is #"100 mL"#, so:

#V_(A^(-)) ~~ 100 - V_(HA)# in units of #"mL"#

Therefore, we now have:

#0.6310 = ("0.240 M" xx (100 - V_(HA)))/("0.100 M" xx V_(HA))#

For ease of notation, let #x = V_(HA)#. Then we have implied units:

#0.6310 = (0.240(100 - x))/(0.100x)#

#= (24.0 - 0.240x)/(0.100x)#

#0.0631x = 24.0 - 0.240x#

#(0.0631 + 0.240)x = 24.0#

#=> x = color(blue)(V_(HA)) = (24.0/(0.0631 + 0.240)) "mL"#

#=# #color(blue)("79.18 mL aqueous benzoic acid")#

That means

#color(blue)(V_(A^(-)) = "20.82 mL aqueous sodium benzoate")#.


As a check, let's see if doing a dilution calculation gives the same ratio.

#"0.240 M benzoate" xx ("20.82 mL")/("100.0 mL")#

#=# #"0.04997 M A"^(-)#

#"0.100 M benzoic acid" xx ("79.18 mL")/("100.0 mL")#

#=# #"0.07918 M HA"#

Therefore, the ratio is:

#\frac(["A"^(-)])(["HA"]) = ("0.04997 M A"^(-))/("0.07918 M HA")#

#= 0.6311 ~~ 0.6310# #color(blue)(sqrt"")#

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