Question

You use 25mL of 6.0M NaOH to neutralize 100mL of an unknown acid in a titration experiment. What is the molarity of the acid? Show work.

Answer 1

You use 25mL of 6.0M NaOH to neutralize 100mL of an unknown acid in a titration experiment. What is the molarity of the acid? Show work

Molarity (C) is defined as the amount of a constituent ( $n_i$ ) divided by the volume of the mixture ( $V$ ), so if given molarity of NaOH is 6.0 mol/L and volume is 25 ml, or 0.025L the amount is:

$$\mathrm{n = V * C = 0.025L * 6.0 \, mol/L = 0.15 \, mol}.$$

There are some different cases of finding molarity of the acid and it is connected with number of protons in acid structure:

When number is 1, the amount ratio of NaOH and acid is 1: 1.

$$\mathrm{NaOH + HX \rightarrow NaX + H_2O}$$

Molarity in this case is 0.15 mol per 100 ml, or 1.5 mol/L (1.5M)

When number is 2, the amount ratio of NaOH and acid is 2: 1.

$$2\mathrm{NaOH + H_2X \rightarrow Na_2X + 2H_2O}$$

Molarity in this case is 0.15/2 mol per 100 ml, or 0.75mol/L (0.75M)

When number is n, the amount ratio of NaOH and acid is n: 1.

Molarity in this case is 0.15/n mol per 100 ml, or ((0.15/n * 100 ml/1000 ml)M)