A beaker with #1.50 xx 10^2"mL"# of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.00 mL of a 0.360 M HCl solution to the beaker?

How much will the pH change? The pKa of acetic acid is 4.740.

1 Answer
Mar 22, 2017

Answer:

#Delta_"pH" = 0.299#

Explanation:

!! VERY LONG ANSWER !!

The first thing to do here is to figure out the concentrations of acetic acid and of acetate anions present in the buffer.

You know that a weak acid/conjugate base buffer has

#color(blue)(ul(color(black)("pH" = "p"K_a + log( (["conjugate base"])/(["weak acid"]))))) -># the Henderson - Hasselbalch equation

In your case, you have

#5.000 = 4.740 + log( (["CH"_3"COO"^(-)])/(["CH"_3"COOH"]))#

Even without doing any calculations, you should be able to say that this buffer contains more conjugate base than weak acid. This is the case because the pH of the buffer is greater than the #"p"K_a# of the acid.

You can rewrite the equation as

#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 5.000 - 4.740#

#log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 0.26#

This will be equivalent to

#10^log((["CH"_3"COO"^(-)])/(["CH"_3"COOH"])) = 10^0.26#

#(["CH"_3"COO"^(-)])/(["CH"_3"COOH"]) = 1.8197#

This implies that

#["CH"_3"COO"^(-)] = 1.8197 * ["CH"_3"COOH"]#

You also know that the total molarity of weak acid and conjugate base is equal to #"0.100 M"#, so you can say that

#["CH"_3"COOH"] + ["CH"_3"COO"^(-)] = "0.100 M"#

You now have two equations with two unknowns. For simplicity, I'll use

#["CH"_3"COOH"] = x" "# and #" "["CH"_3"COO"^(-)] = y#

You have

#x =0.100 - y#

Plug this into the fist equation to get

#y = 1.8197 * (0.100 - y)#

Rearrange to get

#2.8197 * y = 0.18197#

This will be equivalent to

#y = 0.18197/2.8197 = 0.06454#

This means that

#x = 0.100 - 0.06454 = 0.03546#

Therefore, you can say that the buffer contains

#["CH"_3"COOH"] = "0.03546 M"" "# and #" " ["CH"_3"COO"^(-)] = "0.06454 M"#

As predicted, the buffer contains more conjugate base than weak acid.

Now, you are adding

#7.00 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.360 moles H"_3"O"^(+))/(1color(red)(cancel(color(black)("L"))))="0.002520 moles H"_3"O"^(+)#

to the buffer by way of the hydrochloric acid solution. The hydronium cations will react with the acetate anions to produce acetic acid and water

#"CH"_ 3"COO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> "CH"_ 3"COOH"_ ((aq)) + "H"_ 2"O"_ ((l))#

Use the molarity of the acetate anions and the volume of the buffer to calculate the number of moles of conjugate base and of weak acid present in the buffer

#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.06454 moles CH"_3"COO"^(-))/(1color(red)(cancel(color(black)("L"))))#

# = "0.009681 moles CH"_3"COO"^(-)#

#1.50 * 10^2 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.03546 moles CH"_3"COOH")/(1color(red)(cancel(color(black)("L"))))#

# = "0.005319 moles CH"_3"COOH"#

The reaction will consume hydronium cations and acetate anions in a #1:1# mole ratio and produce acetic acid in a #1:1# mole ratio, so you can say that after the neutralization is complete, the solution will contain

#n_ ("H"_ 3"O"^(+)) = "0 moles" -># completely consumed

#n_ ("CH"_ 3"COO"^(-)) = "0.009681 moles" - "0.002520 moles"#

# = "0.007161 moles CH"_3"COO"^(-)#

#n_( "CH"_3 "COOH") = "0.005319 moles" + "0.002520 moles"#

# = "0.007839 moles CH"_3"COOH"#

The total volume of the solution will be

#V_"total" = 1.50 * 10^2"mL" + "7.00 mL" = "157 mL"#

The new concentrations of the weak acid and of the conjugate base will be

#["CH"_3"COOH"] = "0.007839 moles"/(157 * 10^(-3)"L") = "0.04993 M"#

#["CH"_3"COO"^(-)] = "0.007161 moles"/(157 * 10^(-3)"L") = "0.04561M"#

Finally, use the Henderson - Hasselbalch equation to calculate the new pH of the solution

#"pH" = 4.740 + log( (0.04561 color(red)(cancel(color(black)("M"))))/(0.04993color(red)(cancel(color(black)("M")))))#

#"pH" = 4.701#

Therefore, the pH of the solution changed by

#Delta_"pH" = |4.701 - 5.000| = color(darkgreen)(ul(color(black)(0.299)))#

The answer is rounded to three decimal places, the number of significant figures you have for your volumes and molarities.

So, does the result make sense?

You added hydrochloric acid, a strong acid, to the buffer, so you should expect its pH to decrease as a result.

Notice that the final concentration of the weak acid is greater than that of the weak base, which is why you have

#4.701 < 4.740" "# or #" ""pH" < "p"K_a#