Question

Consider the titration of a 26.0-mL sample of 0.185 MCH 3 ​ NH 2 ​ (K b ​ =4.4×10 −4 ) with 0.155M HBr.

Answer 1

The pH of the solution after the titration of a 26.0-mL sample of 0.185 M
`\(CH_3NH_2\) (\(K_b = 4.4 * 10^(-4)\))` with 0.155 M HBr will be acidic.

Step-by-step explanation

Upon titration, the acidic HBr reacts with the basic
`\(CH_3NH_2\).` The equilibrium expression for
`\(CH_3NH_2\)` in water is
`\(CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^+ + OH^-\)`. Since
`\(K_b\)` is given, we can calculate
`\(OH^-\)` concentration at equilibrium. Initially, there is no
`\(OH^-\)` present. The reaction consumes HBr and produces
`\(OH^-\)` ions, leading to an increase in
`\(OH^-\)` concentration. The
`\(OH^-\)` ions make the solution basic, but as the concentration of HBr is higher, the resulting solution will be acidic overall.

Mathematically, we can set up an ICE table (Initial, Change, Equilibrium) to track the concentrations. Initially,
`\(OH^-\)` is 0, and
`\(CH_3NH_2\)` is at its initial concentration. The change is determined by the stoichiometry of the reaction, and at equilibrium, the
`\(OH^-\)` concentration can be calculated. With the
`\(K_b\)` value and
`\(OH^-\)` concentration, we can find the pOH and then the pH of the solution.

In conclusion, the titration of
`\(CH_3NH_2\)` with HBr results in an acidic solution due to the excess of the acidic HBr over the basic
`\(CH_3NH_2\)`. This understanding is crucial for predicting the pH of the solution and evaluating the chemical behavior during the titration process.

Answer 2

The volume of
`\(0.155 \, \text{M} \, \text{HBr}\)` required to neutralize the
`\(0.185 \, \text{M} \, \text{CH}_3\text{NH}_2\)` solution is approximately
`\(0.031 \, \text{L}\) or \(31.0 \, \text{mL}\)`.

To analyze the titration of a 26.0 mL sample of
`\(0.185 \, \text{M} \, \text{CH}_3\text{NH}_2\)` (methylamine, a weak base) with
`\(0.155 \, \text{M} \, \text{HBr}\)` (a strong acid), we'll follow these steps:

1. Calculate Moles of CH3NH2 :

`\[ \text{Moles of CH}_3\text{NH}_2 = \text{Volume} * \text{Concentration} \]`

`\[ = 0.026 \, \text{L} * 0.185 \, \text{M} \]`

2. Calculate Moles of HBr Needed for Neutralization**:

The reaction between
`\( \text{CH}_3\text{NH}_2 \)` and
`\( \text{HBr} \)` is 1:1. So, the moles of
`\( \text{HBr} \)` needed will be equal to the moles of
`\( \text{CH}_3\text{NH}_2 \).`

3. Calculate Volume of HBr Required :

`\[ \text{Volume of HBr} = \frac{\text{Moles of CH}_3\text{NH}_2}{\text{Concentration of HBr}} \]`

This will give us the volume of
`\(0.155 \, \text{M} \, \text{HBr}\)` required to neutralize the
`\(0.185 \, \text{M} \, \text{CH}_3\text{NH}_2\)` solution. Let's perform these calculations.

The volume of
`\(0.155 \, \text{M} \, \text{HBr}\)` required to neutralize the
`\(0.185 \, \text{M} \, \text{CH}_3\text{NH}_2\)` solution is approximately
`\(0.031 \, \text{L}\) or \(31.0 \, \text{mL}\)`. This is the amount of acid needed to reach the equivalence point in the titration.