Chemistry Titration Problem
Below is the experimental data submitted by a student in a chemistry lab. She was given a titrant of oxalic acid whose normality was N/10 (f=0.99).
Experimental Data
| S.N. | Volume of OXALIC ACID | Initial reading of $$KMnO_4$$ | Final reading of $$KMnO_4$$ |
|---|---|---|---|
| 1 | 10 ml | 0 | 10.4 |
| 2 | 10 ml | 10.4 | 22 |
| 3 | 10 ml | 22 | 32.4 |
| 4 | 10 ml | 32.4 | 42.4 |
Questions
- What will be the strength of potassium permanganate in gram per liter?
- Bench acid is used in this titration along with oxalic acid. Can you use HCl instead?
- A teacher asked the topper from the group, Bismita, to titrate bench acid with 0.1 N $$Na_2CO_3$$. She was suggested to perform a rough titration at first. What is the significance of doing that?
- On rough estimation, bench acid was found to have a strength of around 2.4 N. If Bismita was to prepare a diluted acid solution in a 100 ml flask, what would be the extent of dilution of the acid?
- Using the same extent of dilution, if Bismita got a concurrent reading at 12.8 ml when titrating with 10 ml of $$Na_2CO_3$$, what was the actual concentration of bench acid?
Solution
The balanced chemical reaction between potassium permanganate ($$KMnO_4$$) and oxalic acid ($$H_2C_2O_4$$) in an acidic medium is as follows:
a. Strength of potassium permanganate
To determine the concordant volume, we calculate the volume used for each reading:
- Reading 1: $$10.4 - 0 = 10.4 \text{ ml}$$
- Reading 2: $$22 - 10.4 = 11.6 \text{ ml}$$
- Reading 3: $$32.4 - 22 = 10.4 \text{ ml}$$
- Reading 4: $$42.4 - 32.4 = 10.0 \text{ ml}$$
Using the average of the most consistent readings (10.4, 10.4, and 10.0 ml), the average volume of $$KMnO_4$$ is $$10.27 \text{ ml}$$.
Using the normality equation $$N_{oxalic}V_{oxalic} = N_{KMnO_4}V_{KMnO_4}$$:
$$(0.1 \times 0.99 \text{ N}) \times (10 \text{ ml}) = N_{KMnO_4} \times (10.27 \text{ ml})$$
The normality of $$KMnO_4$$ is $$N_{KMnO_4} = 0.0964 \text{ N}$$.
Strength in g/L = Normality x Equivalent Weight. The equivalent weight of $$KMnO_4$$ is $$158.04/5 = 31.61 \text{ g/eq}$$.
Strength = $$0.0964 \text{ N} \times 31.61 \text{ g/eq} = 3.046 \text{ g/L}$$
Answer: The strength of potassium permanganate is approximately 3.05 g/L.
b. Using HCl
Answer: No, HCl cannot be used.
HCl would be oxidized by the strong oxidizing agent $$KMnO_4$$ to form chlorine gas ($$Cl_2$$), which would interfere with the primary titration reaction and lead to inaccurate results.
c. Significance of a rough titration
Answer: To find the approximate volume of titrant needed to reach the endpoint.
This allows subsequent, more accurate titrations to be performed efficiently by adding the titrant quickly at first and then drop-wise near the endpoint, preventing overshooting.
d. Extent of dilution
Dilution factor = Initial Normality / Target Normality = $$2.4 \text{ N} / 0.1 \text{ N} = 24$$
Using $$C_1V_1 = C_2V_2$$ for a 100 ml flask:
$$(2.4 \text{ N}) \times V_1 = (0.1 \text{ N}) \times (100 \text{ ml})$$
$$V_1 = 4.17 \text{ ml}$$
Extent of dilution = Final Volume / Initial Volume = $$100 \text{ ml} / 4.17 \text{ ml} \approx 24$$
Answer: The extent of dilution is a factor of approximately 24.
e. Actual concentration of bench acid
Using the normality equation for the diluted acid and $$Na_2CO_3$$:
$$N_{acid} \times (12.8 \text{ ml}) = (0.1 \text{ N}) \times (10 \text{ ml})$$
$$N_{acid} = 0.078125 \text{ N}$$
Actual Concentration = Normality of diluted acid x Dilution Factor
Actual Concentration = $$0.078125 \text{ N} \times 24 = 1.875 \text{ N}$$
Answer: The actual concentration of the bench acid is approximately 1.88 N.