Question

A 0.4159-g sample of primary-standard was treated with 42.00 mL of dilute perchloric acid. The solution was boiled to remove , following which the excess was back-titrated with 9.65 mL of dilute . In a separate experiment, it was established that 26.76 mL of the neutralized the in a 25.50-mL portion. Calculate the molarities of the and .

Answer 1

The concentration of the original hydrochloric acid solution sample is calculated using the volume and molarity of the sodium hydroxide used in the titration, resulting in a concentration of 0.019385 M HCl.

Step-by-step explanation:

To find the concentration of hydrochloric acid (HCl), we use the volume and molarity of the sodium hydroxide (NaOH) used in the titration. In this case, 31.416 mL or 0.031416 liters of 0.1234 M NaOH is used to neutralize the HCl in the solution. The balanced chemical equation for the reaction is:

HCl + NaOH → NaCl + H2O

Since the molar ratio of HCl to NaOH is 1:1, the moles of HCl neutralized will be equal to the moles of NaOH used. We calculate the moles of NaOH and, hence, moles of HCl:

Number of moles of NaOH = Molarity × Volume = 0.1234 M × 0.031416 L = 0.003877 mol

Assuming the sample of HCl was 200.0 mL or 0.200 L, we can then calculate the molarity of HCl:

Molarity of HCl = Number of moles of HCl / Volume of HCl = 0.003877 mol / 0.200 L = 0.019385 M