Question

Calculate the molarity of a sodium hydroxide solution if 24.3 mL were required to titrate 25.00 mL of a 0.873 M acetic acid solution.

Answer 1

Answer on Question #63184 – Chemistry – Other

Task:

Calculate the molarity of a sodium hydroxide solution if 24.3 mL were required to titrate 25.00 mL of a 0.873 M acetic acid solution.

Solution:

The reaction between acetic acid, CH₃COOH, and sodium hydroxide, NaOH, is shown below:

$$CH_3COOH + NaOH = CH_3COONa + H_2O$$

At the equivalence point the moles of the acid (CH₃COOH) are equal to the moles of the base (NaOH). We can use the known concentration and measured volume of the acetic acid to find the number of moles of acid used in the titration:

$$\text{moles } CH_3COOH = \frac{0.873 \text{ mol } CH_3COOH}{1 \text{ L}} \times 0.025 \text{ L } CH_3COOH = 0.021825 \text{ mol } CH_3COOH$$

At the equivalence point: mol NaOH = mol CH₃COOH, so moles of NaOH = 0.021825 mol.

The concentration of the sodium hydroxide is equal to the number of moles divided by its volume:

$$\text{molarity of NaOH} = C(\text{NaOH}) = \frac{0.021825 \text{ mol NaOH}}{0.0243 \text{ L}} = 0.898 \text{ M}$$

Answer: $C(\text{NaOH}) = 0.898 \text{ M}$

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