Question

A 25.0 ml sample of 0.150 m benzoic acid is titrated with a 0.150 m naoh solution. what is the ph at the equivalence point? the ka of benzoic acid is 4.50 × 10-4.

Answer 1

pH = 8.11

Step-by-step explanation:

Step 1: Data given

Volume of a 0.150 M benzoic acid sample = 25.0 mL

Molarity of NaOH solution = 0.150 M

Ka of the benzoic acid = 4.50 * 10^-4

Step 2: The balanced equation

C7H6O2 + NaOH → C7H5O2Na + H2O

Step 3: calculate moles

At the equivalence point we have the same amount of of acid and base

Both benzoic acid and NaOH will completely be consumed.

Moles benzoic acid = molarity *volume

Moles benzoic acid = 0.150 M * 0.025 L

Moles benzoic acid = 0.00375 moles

We we'll need 0.00375 moles NaOH to reach the equivalence point

Volume NaOH = 0.00375 moles / 0.150 M

Volume = 0.025 L = 25 mL

Total volume = 25 + 25 = 50 mL

Step 4: Calculate moles of C7H5O2Na

For 1 mol benzoic acid we need 1 mol NaOH to produce 1 mol C7H5O2Na

For 0.00375 moles benzoic acid we'll have 0.00375 moles C7H5O2Na

Step 5: The equation

C7H5O2- + H2O → C7H6O2 + OH-

Step 6: The initial concentration

[C7H5O2-] = 0.00375 moles / 0.050 L

[C7H5O2-] = 0.075 M

[C7H6O2] = 0M

[OH-] = 0M

Step 7: The concentration at the equilibrium

[C7H5O2-] = 0.075 - X M

[C7H6O2] = XM

[OH-] = XM

Step 8: Calculate Kb

Ka * Kb = Kw

Kb = Kw / Ka

Kb = 10^-14 / (4.50 *10^-4)

Kb = 2.22 *10^-11

Kb = [C7H6O2][OH-]/[C7H5O2-]

2.22 * 10^-11 = X²/ 0.075 - X

2.22 * 10^-11 = X²/ 0.075

X² = 2.22 *10^-11 * 0.075

X² = 1.665 *10^-12

X = 1.29 * 10^-6 = [OH-]

Step 9: Calculate pOH

pOH = -log[OH-]

pOH = -log(1.29 * 10^-6 )

pOH = 5.89

Step 10: Calculate pH

pH = 14 - 5.89

pH = 8.11

Answer 2

At the equivalence point, moles NaOH = moles benzoic acid

HA + NaOH ==> NaA + H2O where HA is benzoic acid

At the equivalence point, all the benzoic acid ==> sodium benzoate

A^- + H2O ==> HA + OH- (again, A^- is the benzoate anion and HA is the weak acid benzoic acid)

Kb for benzoate = 1x10^-14/4.5x10^-4 = 2.22x10^-11

Kb = 2.22x10^-11 = [HA][OH-][A^-] = (x)(x)/0.150

x^2 = 3.33x10^-12

x = 1.8x10^-6 = [OH-]

pOH = -log [OH-] = 5.74

pH = 14 - pOH = 8.26