# 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?

##### 1 Answer

#### Answer

#### Answer:

#### Explanation

#### Explanation:

Well, we know the final **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; **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

#=> 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^(-)) ~~ 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

#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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