Question #f873f

!! VERY LONG ANSWER !!

Start by writing the balanced chemical equation that describes this neutralization reaction

#"C"_ 6"H"_ 5"COOH"_ ((aq)) + "NaOH"_ ((aq)) -> "C"_ 6"H"_ 5"COONa"_ ((aq)) + "H"_ 2"O"_ ((l))#

Notice that the two reactants react in a #1:1# mole ratio, so right from the start, you know that the number of moles of strong base must be equal to the number of moles of weak acid present in solution.

Convert the mass of benzoic acid to moles by using the compound's molar mass

#0.976 color(red)(cancel(color(black)("g"))) * ("1 mole C"_6"H"_5"COOH")/(122.12color(red)(cancel(color(black)("g")))) = "0.007992 moles C"_6"H"_5"COOH"#

This means that the sodium hydroxide solution must contain #0.007992# moles of sodium hydroxide. Use its molarity to calculate the volume that would contain that many moles of solute

#0.007992 color(red)(cancel(color(black)("moles NaOH"))) * "1 L solution"/(0.100 color(red)(cancel(color(black)("moles NaOH")))) = "0.07992 L"#

Expressed in milliliters and rounded to three sig figs, the answer will be

#color(darkgreen)(ul(color(black)(V_"NaOH" = "79.9 mL")))#

Now, after the reaction is complete, the resulting solution will contain aqueous sodium benzoate, the salt of the benzoate anion, #"C"_6"H"_5"COO"^(-)#, the conjugate base of benzoic acid.

The benzoate anion will react with water to reform some of the benzoic acid and produce hydroxide anions, so right from the start, you should expect the pH of the solution to be #> 7#, i.e. the resulting solution will be slightly basic.

The reaction consumed #0.007992# moles of weak acid and of strong base and produced #0.007992# moles of benzoate anions in a volume of #"79.9 mL"# of solution.

The concentration of the benzoate anions will be

#["C"_6"H"_5"COO"^(-)] = "0.007992 moles"/(79.9 * 10^(-3)"L") = "0.100 M"#

You must assume that the volume of the solution is equal to the volume of added sodium hydroxide solution

Use an ICE table to find the concentration of the hydroxide anions in the resulting solution

#"C"_ 6"H"_ 5"COO"_ ((aq))^(-) + "H"_ 2"O" _ ((l)) rightleftharpoons "C"_ 6"H"_ 5"COOH"_ ((aq)) + "OH"_ ((aq))^(-)#

#color(purple)("I")color(white)(aaaacolor(black)(0.100 )aaaaaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaacolor(black)(0)aaa#
#color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaaaaaaaaacolor(black)((+x))aaaaaaaacolor(black)((+x))aaa#
#color(purple)("E")color(white)(aacolor(black)(0.100 -x)aaaaaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaaaaaacolor(black)(x)aaa#

By definition, the base dissociation constant, #K_b#, will take the form

#K_b = (["C"_6"H"_5"COOH"] * ["OH"^(-)])/(["C"_6"H"_5"COO"^(-)])#

Keep in mind that for an aqueous solution at #25^@"C"#, you have

#color(blue)(ul(color(black)(K_a * K_b = 10^(-14))))#

This means that the base dissociation constant for the benzoate anion will be

#K_b = 10^(-14)/(6.5 * 10^(-5)) = 1.54 * 10^(-10)#

Therefore, you can say that

#1.54 * 10^(-10) = (x * x)/(0.100 -x ) = x^2/(0.100 - x)#

Since #K_b# has such a very small value compared with the concentration of the benzoate anion, you can use the approximation

#0.100 - x ~~ 0.100#

This means that you will have

#1.54 * 10^(-10) = x^2/0.100#

which will get you

#x = sqrt(0.100 * 1.54 * 10^(-10)) = 3.924 * 10^(-6)#

Since #x# represents the equilibrium concentration of hydroxide anions, you can say that you will have

#["OH"^(-)] = 3.924 * 10^(-6)"M"#

Use the fact that

#color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))#

to calculate the pOH of the solution

#"pOH" = - log(3.924 * 10^(-6)) = 5.406#

As you know, an aqueous solution at #25^@"C"# has

#color(blue)(ul(color(black)("pH " + " pOH" = 14)))#

which means that the pH of this solution will be

#color(darkgreen)(ul(color(black)("pH" = 14 - 5.406 = 8.594)))#

The answer is rounded to three decimal points, the number of significant figures you have for your values.

As predicted, the resulting solution is indeed slightly basic.