Question

For each trial with the unknown acid, calculate the following:

2. For each trial with the unknown acid, calculate the following Trial 2 a. Moles of NaOH required to reach the equivalence point b. Moles of unknown acid in the beaker c. Molar mass of unknown acid Trial 3 a. Moles of NaOH required to reach the equivalence point b. Moles of unknown acid in the beaker c. Molar mass of unknown acid

Here is my data:

Answer 1

The unknown acid is monoprotic as there is only one equivalence point. Let the acid be HA and then its reaction with NaOH will be

$HA(aq)+NaOH(aq)\rightarrow NaA(aq)+H_{2}O(l)$   (1)

Using the average concentration of the NaOH in the calculations. [NaOH]_avg = 0.0971 M

Trial 2 :

Initial volume of NaOH : V_i = 11.3 mL

Volume of NaOH at equivalence point : V_f = 22.7 mL

Thus, the volume of NaOH required to neutralize the acid is

$V_{NaOH}=V_{f}-V_{i}=11.4 mL$   (2)

a. So moles of NaOH (n_NaOH) required to reach equivalence point is

$n_{NaOH}=V_{NaOH}*[NaOH]_{avg}=11.4mL*\frac{0.0971mol}{1000mL}=0.0011mol$   (3)

b. Using the stoichiometry of the reaction given in equation (1), 1 mol NaOH has reacted with 1 mol acid at the equivalence point.

So, the moles of unknown acid in the beaker is

$n_{HA}=n_{NaOH}=0.0011mol$   (4)

c. Molar mass of unknown acid is given by

$MM_{HA}=\frac{m_{HA}}{n_{HA}}=\frac{0.204g}{0.0011mol}=185.45gmol^{-1}$   (5)

From the pH at the equivalence point, we see that the pH > 7, i.e. this is a titration of a weak acid (HA) with strong base (NaOH) where the conjugate base of the weak acid (A^-) will react with water to generate hydroxide ions thus making the solution at the equivalence point basic. (A^- + H₂O -> HA + OH^-).

Similarly, for Trial 3 :

a. $n_{NaOH}=V_{NaOH}*[NaOH]_{avg}=(39-28)mL*\frac{0.0971mol}{1000mL}=1.07mmol$    (6)

b. At the equivalence point, moles of acid = moles of base added, thus,

$n_{HA}=n_{NaOH}=1.07mmol$   (7)

c. Molar mass of unknown acid is then given by

$MM_{HA}=\frac{m_{HA}}{n_{HA}}=\frac{0.206g}{1.07mmol}=\frac{0.206g}{1.07*10^{-3}mol}=192.52gmol^{-1}$ (8)