Chemistry 2002 JAMB Past Questions - Question 19: Three drops of a 1.0mol dm³ solution of NaOH are added to 20cm³ of a solution of pH 8.4. The pH of the resulting solution will be

Correct Answer

C

Explanation

To determine the pH of the resulting solution after adding three drops of a 1.0 mol/dm³ solution of NaOH to 20 cm³ of a solution with pH 8.4, we need to consider the neutralization reaction between NaOH and H⁺ ions in the solution.

The balanced chemical equation for the reaction between NaOH and H⁺ is:

\[ \text{NaOH} + \text{H}^+ \rightarrow \text{Na}^+ + \text{H}_2\text{O} \]

In this case, three drops of a 1.0 mol/dm³ solution of NaOH are added. The concentration of OH⁻ ions in the solution after the addition can be calculated as follows:

\[ \text{Moles of NaOH added} = \text{concentration} \times \text{volume} \]
\[ \text{Moles of NaOH} = 1.0 \, \text{mol/dm}³ \times (3 \, \text{drops} \times \text{drop volume}) \]

Assuming one drop is approximately 0.05 cm³, the volume of NaOH added is \(3 \times 0.05 \, \text{cm}³ = 0.15 \, \text{cm}³\).

\[ \text{Moles of NaOH} = 1.0 \, \text{mol/dm}³ \times 0.15 \, \text{dm}³ \]

Now, we need to consider the reaction with H⁺ ions. One mole of NaOH reacts with one mole of H⁺ ions. So, the moles of H⁺ ions neutralized by NaOH is also \(0.15 \, \text{mol}\).

The initial moles of H⁺ ions can be calculated using the formula:

\[ \text{Moles of H}^+ = 10^{-\text{pH}} \times \text{volume} \]

Given that the initial pH is 8.4 and the volume is 20 cm³:

\[ \text{Moles of H}^+ = 10^{-8.4} \times 0.02 \]

Now, we subtract the moles of H⁺ neutralized by NaOH from the initial moles of H⁺ to find the remaining moles of H⁺:

\[ \text{Remaining moles of H}^+ = \text{Initial moles of H}^+ - \text{Moles of H}^+ \]

Finally, we can calculate the pH of the resulting solution using the formula:

\[ \text{pH} = -\log(\text{H}^+) \]

Plug in the remaining moles of H⁺ into this formula to find the pH of the resulting solution.