Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough w

Question

Assume you dissolve 0.235 g of the weak acid benzoic acid, C6H5CO2H, in enough water to make 3.00 102 mL of solution and then titrate the solution with 0.133 M NaOH. C6H5CO2H(aq) + OH-(aq) C6H5CO2-(aq) + H2O() What are the concentrations of the following ions at the equivalence point? Na+, H3O+, OH- C6H5CO2- Incorrect: Your answer is incorrect. M Na+ Incorrect: Your answer is incorrect. M H3O+ Incorrect: Your answer is incorrect. M OH- Incorrect: Your answer is incorrect. M C6H5CO2- What is the pH of the solution? Incorrect: Your answer is incorrect.

Explanation / Answer

First, let's calculate the moles of the acid:

moles A = 0.235 g / 122.12 g/mol = 0.00192 moles

With the expression moles A = moles B we can calculate the volume at the equivalence point:
0.00192 = 0.133 * Vb
Vb = 0.00192 / 0.133 = 0.01447 L or 14.47 mL.

At the equivalence point, all the acid is being consumed and the base begins to be in excess, so, the predominant form will be the conjugate base of the acid at the final volume:

Total volume = 300 + 14.47 = 314.47 mL or 0.31447 L
moles of Base = 0.00192 moles
Innitial concentration of the conjugate base = 0.00192 / 0.31447 = 0.0061 M

The overall reaction:
r: C6H5COO- + H2O <------> C6H5COOH + OH-
i. 0.0061 0 0
e. 0.0061-x x x

The Kb for this is: 1.6x10-10 so
1.6x10-10 = x2 / 0.0061-x --> Ka is small so x will be small too, and we can neglect the sum so:
1.6x10-10(0.0061) = x2
x = [OH-] = [C6H5COOH] = 9.88x10-7 M
[C6H5COO-] = [Na+] = 0.0061 M

To get the pH, and H3O+ let's calculate pOH first:
pOH = -log(9.88x10-7) = 6.01
pH = 14-6.01 = 7.99
[H3O+] = 10-7.99 = 1.02x10-8 M

Hope this helps

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