Question #e8a1b

You are correct, the pH of the buffer solution that you prepared in part (A) is equal to #5.13#.

Now, how would you expect the pH of the solution in part (B) to compare with that of the first solution?

What you're essentially doing here is taking a sample of the first solution and diluting it by adding water. So would that have an impact on the pH of the solution?

Simply put, no. You can expect the pH of the second solution to be equal to that of the first solution. Here's why that happens.

You calculated the number of moles of acetic acid, #n_1#, and acetate anions, #n_2#, present in the first solution to be

#color(blue)(c = n/V implies n = c * V)#

#n_1 = "1.0 M" * 10.0 * 10^(-3)"L" = "0.0100 moles CH"_3"COOH"#

#n_2 = "1.0 M" * 24.0 * 10^(-3)"L" = "0.0240 moles CH"_3"COO"^(-)#

Now, your #"10-mL"# sample will contain - keep in mind that the total volume of the first solution will be equal to #"34 mL"#

#10.0color(red)(cancel(color(black)("mL solution"))) * ("0.0100 moles CH"_3"COOH")/(34color(red)(cancel(color(black)("mL solution")))) = "0.0029412 moles CH"_3"COOH"#

and

#10.0color(red)(cancel(color(black)("mL solution"))) * ("0.0240 moles CH"_3"COO"^(-))/(34color(red)(cancel(color(black)("mL solution")))) = "0.0070588 moles CH"_3"COO"^(-)#

When you dilute this solution, you keep the number of moles of the two species constant, but you're increasing the volume of the solution.

#V_"second solution" = "10.0 mL" + "40.0 mL" = "50.0 mL"#

The concentrations of the two species will now be

#["CH"_3"COOH"] = "0.0029412 moles"/(50.0 * 10^(-3)"L") = "0.058824 M"#

#["CH"_3"COO"^(-)] = "0.0070588 moles"/(50.0 * 10^(-3)"L") = "0.14118 M"#

Plug these values into the Henderson - Hasselbalch equation - the #pK_a# of acetic acid is #4.75#

#color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))#

#"pH" = 4.75 + log( (0.14118color(red)(cancel(color(black)("M"))))/(0.058824color(red)(cancel(color(black)("M"))))) = 5.13#

The pH of a buffer solution is determined by the ratio that exists between the concentrations of the weak acid and of its conjugate base.

More specifically, since the volume is the same for both, the pH is determined by the mole ratio that exists between the two species.

Taking the #"10-mL"# sample preserves this mole ratio, so diluting the sample will have no impact on the pH, even though the actual concentrations change.