A buffer that contains 1.10 mol/L B and 0.700 mol/L BH+ has a pH of 9.65. What is the pH after 0.0045 mol of HCl is added to 0.500 L of this solution?

1 Answer
Truong-Son N.
Apr 5, 2018

If we have a buffer, then the Henderson-Hasselbalch equation applies:

#"pH" = "pK"_a + log\frac(["B"])(["BH"^(+)])#

We evidently do not know the #"pK"_a#, so let's solve for it first.

#"pK"_a = "pH" - log\frac(["B"])(["BH"^(+)])#

#= 9.65 - log("1.10 M B"/"0.700 M BH"^+)#

#= 9.45#

This makes sense since #"pH" > "pK"_a# when there is more weak base than weak acid. In #"0.500 L"# of solution total, we can then find the mols present of both.

#"1.10 M B" xx "0.500 L" = "0.550 mols B"#

#"0.700 M BH"^(+) xx "0.500 L" = "0.350 mols BH"^(+)#

The #"0.0045 mols HCl"# then reacts with the weak base and generates weak acid, i.e. consumes #"B"# and makes #"BH"^(+)# in an equimolar fashion. Thus,

#"pH"' = 9.45 + log(("0.550 mols B" - "0.0045 mols HCl")/("0.350 mols BH"^(+) + "0.0045 mols HCl"))#

The #"pH"# should go down... I might expect something close to #9.6#, but it should be lower than #9.65#.

#color(blue)("pH"') = 9.45 + 0.187#

#= color(blue)(9.64)#

So this is a pretty good buffer. How would it respond to #"0.0045 mols NaOH"#? Did you get #"pH"'' = 9.66#?

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