1.a)How many milliliters of 0.0998 M NaOH are required to “neutralize” (react co

Question

1.a)How many milliliters of 0.0998 M NaOH are required to “neutralize” (react completely) with 0.305 g of benzoic acid, C6H5COOH, (Molar mass = 122.12g/mol).

b)What is the pH of the resulting solution after the “neutralization” reaction (that is after the NaOH solution has been added to the benzoic acid)? pKa of benzoic acid is 4.20. (You may assume the solid benzoic acid sample was dissolved in 25.0 mL of distilled, deionized water before being reacted with the base to assist in determining the volume of the solution after the reaction is complete.)

Explanation / Answer

1a) Benzoic acid moles = Mass/Molar mass = ( 0.305 /122.12) = 0.0025

Moles of NaOH for neutralisation = benzoic acid moles = 0.0025

Moles of NaoH =M x V = 0.0998 x V = 0.0025 , V = 0.025 L = 25 ml

hence 25 ml NaOH required

b) solution vol = 25 ml , NaOH vol = 25 ml , total voume = 50 ml = 0.05 L

after neutralisation we have benzoate , benzoate moles = NaOH moles reacted = 0.025

Now we have equilibrium

C6H5COO- + H2O (l) <---> C6H5COOH + OH-

at equilibrium C6H5COO- moles = C6H5COOH = X, = OH- moles ,

Kb of bezate = Kw /Ka of acid = ( 10^ -14) / ( 10^ -4.2)= 1.585 x 10^ -10

Kb = [C6H5COOH] [OH-] /[C6H5COO-]

1.585 x 10^ -10 = ( X/0.05) ( X/0.05) / ( 0.025-X) /0.05

7.92 x 10^ -12 = X^ 2 / ( 0.025-X)

X = 4.45 x 10^ -7

[OH-] = ( 4.45x10^ -7 /0.05) = 8.9 x 10^ -6 M

pOH = -log [OH-] = -log ( 8.9x10^ -6) = 5

pH = 14-5 = 9

1.585 x 10^ -10 =

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