# 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?

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

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

##### 1 Answer

#### 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"]))))) -># theHenderson - 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

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

#["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 **mole ratio** and produce acetic acid in a **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#