Question

4. The student then titrated 25.0 mL of the same buffer with 0.1115 M NaOH. The observed data is below. Hand draw a graph of the data (on graph paper) and determine the concentration of the weak acid in the buffer? Vol of NaOH Added (mL) pH Vol of NaOH Added (mL) pH 0.00 4.91 11.00 10.32 1.00 4.97 12.00 10.99 2.00 5.04 13.00 11.26 3.00 5.11 14.00 11.44 4.00 5.2 15.00 11.57 5.00 5.29 16.00 11.66 6.00 5.4 17.00 11.73 7.00 5.53 18.00 11.79 8.00 5.72 19.00 11.84 9.00 5.99 20.00 11.89 10.00 6.66 21.00 11.91

5. What is the buffer strength from question 3 and 4 (i.e., the total of [HA] and [A- ])? 6. If the pH of the buffer (before it is titrated) is 5.70, what is the pKa of the weak acid?

Answer 1

4: The plot of pH versus Vol.NaOH(mL) is shown below

I have drawn the graph in excel. You can plot the same graph in graph paper.

When we add NaOH, it reacts with the weak acid to form salt of the weak acid and water.

From the above graph it is clear that equivalence point is achieved when 10 mL of NaOH is added.

Veq = 10.00 mL = 0.010 L

Hence moles of 0.1115 M NaOH added till the equilibrium point achieved = MxV = 0.1115 mol/L x 0.010 L

= 0.001115 mol NaOH

Hence 0.001115 mol NaOH will react with 0.001115 mol of the weak acid. i.e

25.0 mL( = 0.0250 L) of the buffer solution contains 0.001115 mol of the weak acid.

Hence concentration of weak acid in the buffer solution, [HA] = 0.001115 mol / 0.0250 L = 0.0446 M (answer)

5: [HA] = 0.0446 M

Q.3 is required to answer the rest.