Question

Analyze the following titration data using Excel, find the endpoint using the intersect method of the provided data, and determine the final molarity if 25.00 mL of an unknown acid is titrated with 0.3097 M NaOH standardized base. {volume NaOH (mL), Conductivity (µS/mL)} = {{0.00, 14654}, {2.00, 14651}, {4.01, 14651}, {6.01, 14648}, {8.01, 14647}, {10.01, 14646}, {12.02, 14644}, {14.74, 14644}, {16.74, 14649}, {18.74, 14653}, {20.75, 14658}, {22.75, 14662}, {24.75, 14666}, {26.75, 14670}}.

Answer 1

NaOH volume mL	Conductivity µS/mL
0	14654
2	14651
4.01	14651
6.01	14648
8.01	14647
10.01	14646
12.02	14644
14.74	14644
16.74	14649
18.74	14653
20.75	14658
22.75	14662
24.75	14666
26.75	14670

14675 14670 14665 14660 14655 14650 14645 14.74 ml NaOH at 14644 conductance 14640 0 10 15 20 25 30 Volume of added NaOH Conductivity

Figure 1. Conductance against Volume of NaOH added.

Before the addition of NaOH, the conductance of solution is high because of the presence of the cations (H⁺). When the NaOH is added, the conductance decrease due to the reaction of cations (H⁺) with hydroxyl ions (OH^-) to form un-dissociated water.

The conductance decrease continuously until the equivalence point. At this point, the solution contains only salt obtained by the reaction of an Acid and a base. After the equivalence point, the conductance gradually increases because of the large conductivity of hydroxyl ions (OH^-). According to Figure 1, the point of intersection of the two lines gives the point of neutralization, which is known as the equivalence point. i.e. 14.74 mL NaOH at 14644 µS/mL conductance.

Calculations for the concentration of unknown acid.

By applying V1S1 = V2S2 equation. Where, V1 = Volume of Unknown acid (25 mL)

S1 = Strength of unknown acid (M)

V2 = Volume of NaOH required for the equivalent point

S2 = Strength of NaOH (0.3094 M)

Therefore, S1 = V2S2/V1

= 14.74 * 0.3097 / 25

= 4.565 / 25

= 0.1826 M acid.

So the molarity of unknown acid is 0.1826 M

From the figure, the curve suggests that the used acid during the conductometry titration is a weak acid, as the value of conductance at the starting point is less than the value of conductance at last point, i.e. at the start, the volume of NaOH is 0.00 and the conductance is 14654 µS/mL, while at last, the volume of NaOH is 26.75 and the conductance is 14670 µS/mL.

And the acid must be Acetic acid.

The reaction is CH₃COOH + NaOH --------> CH₃COONa + H₂O