Question #f0ac6

1 Answer
Apr 23, 2016

Answer:

#"pH" = 4.05#

Explanation:

!! LONG ANSWER !!

You're dealing with a buffer solution that contains benzoic acid, #"C"_7"H"_6"O"_2#, a weak acid, and sodium benzoate, #"C"_7"H"_5"O"_2"Na"#, the salt of its conjugate base, the benzoate anion, #"C"_7"H"_5"O"_2^(-)#.

You can calculate the pH of a buffer by using the Henderson - Hasselbalch equation, which for a buffer that contains a weak acid and its conjugate looks like this

#color(blue)(|bar(ul(color(white)(a/a)"pH" = pK_a + log( (["conjugate base"])/(["weak acid"]))color(white)(a/a)|)))#

Here you have

#color(blue)(|bar(ul(color(white)(a/a)pK_a = - log(K_a)color(white)(a/a)|)))#

Now, before calculating the pH of the buffer after the addition of the hydrochloric acid, try to predict how you expect the pH to change relative to the pH of the original buffer solution.

You're adding a strong acid that will react with the benzoate anions to form benzoic acid. This tells you the addition of the acid will result in a decrease in the concentration of the conjugate base.

The concentration of the weak acid may or may not increase, depending on the change in volume that results from the addition of the hydrochloric acid solution.

As a result, you should expect the pH of the solution to be lower than the pH of the original buffer solution.

So, use the molarity and volume of the hydrochloric acid solution to determine how many moles of acid you're adding to the buffer

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

You will have

#n_(HCl) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(25 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))#

#n_(HCl) = "0.0025 moles HCl"#

Hydrochloric acid will ionize completely in aqueous solution to form hydronium cations, #"H"_3"O"^(+)#, in a #1:1# mole ratio. This means that you have

#n_(H_3O^(+)) = n_(HCl)#

#n_(H_3O^(+)) = "0.0025 moles H"_3"O"^(+)#

The hydronium cations produced by the ionization of the hydrochloric acid will react with the benzoate anions

#"C"_ 7"H"_ 5"O"_ (2(aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> "C"_ 7"H"_ 6"O"_ (2(aq)) + "H"_ 2"O"_((l))#

Notice that #1# mole of hydronium cations will react with #1# mole of benzoate anions and produce #1# mole of benzoic acid. Keep this in mind.

Use the molarities and volumes of the beznoic acid and benzoate anions to get the number of moles of each present in the buffer

#n_"benzoic acid" = "0.25 mol" color(red)(cancel(color(black)("L"^(-1)))) * 0.25 color(red)(cancel(color(black)("L"))) = "0.0625 moles C"_7"H"_6"O"_2#

#n_"benzoate anions" = "0.2 mol" color(red)(cancel(color(black)("L"^(-1)))) * 0.25 color(red)(cancel(color(black)("L"))) = "0.0500 moles C"_7"H"_5"O"_2^(-)#

Now, the hydronium cations will be completely consumed by the reaction with the benzoate anions. Once the reaction is complete, you will have

#n_(H_3O^(+)) = "0 moles" -># completely consumed

#n_"benzoate anions" = "0.0500 moles" - "0.0025 moles"#

#color(white)(a)= "0.0475 moles C"_7"H"_5"O"_2^(-)#

#n_"benzoic acid" = "0.0625 moles" + "0.0025 moles"#

#color(white)(a)= "0.0650 moles C"_7"H"_6"O"_2#

The total volume of the buffer solution will be

#V_"total" = "0.25 L" + 25.0 * 10^(-3)"L" = "0.275 L"#

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

#["C"_7"H"_6"O"_2] = "0.0650 moles"/"0.275 L" = "0.2364 mol L"^(-1)#

#["C"_7"H"_5"O"_2^(-)] = "0.0475 moles"/"0.275 L" = "0.1727 mol L"^(-1)#

As you can see, the concentration of the conjugate base decreased, but so did the concentration of the weak acid. This tells you that the increase in volume prevented the increase in the concentration of the benzoic acid.

The pH of the solution will be

#"pH" = pK_a + log( (["C"_7"H"_5"O"_2^(-)])/(["C"_7"H"_6"O"_2]))#

Plug in your values to find

#"pH" = - log(6.5 * 10^(-5)) + log( (0.1727 color(red)(cancel(color(black)("mol L"^(-1)))))/(0.2364 color(red)(cancel(color(black)("mol L"^(-1))))))#

#"pH" = color(green)(|bar(ul(color(white)(a/a)4.05color(white)(a/a)|)))#

To test the result, calculate the pH of the initial buffer solution. If our calculations are correct, you should get a value that's higher than #4.05#.

But remember, the purpose of a buffer solution is to resist drastic changes in pH when small quantities of strong acids or strong bases are added, so you should expect the pH of the initial buffer solution to be only slightly higher than #4.05#.

#"pH" = - log(6.5 * 10^(-5)) + log( (0.2 color(red)(cancel(color(black)("mol L"^(-1)))))/(0.25 color(red)(cancel(color(black)("mol L"^(-1))))))#

#"pH" = 4.09#

Indeed, the pH of the solution decreased slightly upon the addition of the strong acid.