Question

A 25 mL sample of 0.100 M methylamine is titrated using 0.112 M HCl (aq) at 25

Answer 1

Here, 25 mL of 0.100 M of methylamine is titrated with 15 mL of 0.112 M HCl, it means that only 15 mL of methylamine has reacted therefore, the remaining amount of methylamine = 25 mL - 15 mL = 10mL. And the total volume of the mixture = 25 mL + 15 mL = 40 mL.

The dissociation of methylamine is shown by the following equation,

Molarity of CH₃NH₂ and CH₃NH₃⁺ can be calculated by using the following formula,

Therefore, the [CH₃NH₂] = 0.025 M, [CH₃NH₃⁺] = 0.0375 M and K_b =4.4 * 10^-4 , by substituting these values in equation (i), the concentration of OH^- ion [OH^-] can be calculated as follows,

pOH can be calculated by substituting the value of [OH^-] = 2.93 * 10^-4 M in the following equation,

pH and pOH of a solution are related to each other by the following equation,

Therefore, the pH of the solution can be calculated by substituting the value of pOH = 3.53 in the above equation as follows,

Hence, the pH of this solution is 10.47.