1. State the meaning of the term strong acid - AQA - A-Level Chemistry - Question 5 - 2018 - Paper 1

To calculate the pH of the solution formed, we first need to determine the moles of HCl and Ba(OH)₂:

• Moles of HCl:

$ext{Moles HCl} = 0.100 ext{ mol dm}^{-3} imes 0.01035 ext{ dm}^3 = 0.001035 ext{ mol}$

• Moles of Ba(OH)₂:

$ext{Moles Ba(OH)}_2 = 0.150 ext{ mol dm}^{-3} imes 0.0250 ext{ dm}^3 = 0.00375 ext{ mol}$

Ba(OH)₂ produces 2 moles of OH⁻ ions per mole, so:

$ext{Moles OH}^- = 0.00375 ext{ mol} imes 2 = 0.00750 ext{ mol}$

Now we calculate the total volume of the solution:

$ext{Total Volume} = 10.35 ext{ cm}^3 + 25.0 ext{ cm}^3 = 35.35 ext{ cm}^3 = 0.03535 ext{ dm}^3$

Concentration of OH⁻ ions:

[ ext{OH}^-] = rac{0.00750 ext{ mol}}{0.03535 ext{ dm}^3} = 0.212 ext{ mol dm}^{-3}

Using the ion product constant of water, we can find [H⁺]:

$K_w = [H^+][OH^-]$ $1.47 imes 10^{-14} = [H^+][0.212]$

Solving for [H⁺]:

[H^+] = rac{1.47 imes 10^{-14}}{0.212} = 6.93 imes 10^{-14} ext{ mol dm}^{-3}

Now calculate pH:

$ext{pH} = - ext{log} [H^+] = - ext{log} (6.93 imes 10^{-14}) = 13.13$

Water is neutral at 30°C because the concentration of hydrogen ions [H⁺] is equal to the concentration of hydroxide ions [OH⁻] at this temperature, maintaining a balance between them.

The oxide is SO₂.

Kₐ = \frac{[H^+][CH₃COO^-]}{[CH₃COOH]}

To calculate the pH after adding HCl to the buffer:

Firstly, calculate moles of the components:

• Moles of sodium ethanoate:

$0.025 ext{ mol}$

• Moles of ethanoic acid:

$= 0.700 ext{ mol dm}^{-3} imes 0.500 ext{ dm}^3 = 0.350 ext{ mol}$

Adding HCl:

• Moles of HCl = $2.00 ext{ mol dm}^{-3} imes 0.005 ext{ dm}^3 = 0.010 ext{ mol}$

After addition:

• Moles of ethanoic acid = 0.350 + 0.010 = 0.360 mol
• Moles of sodium ethanoate = 0.025 - 0.010 = 0.015 mol

Using the Henderson-Hasselbalch equation:

$ext{pH} = pK_a + ext{log} \left( \frac{[A^-]}{[HA]} \right)$

Substituting:

• Given Kₐ = $1.76 \times 10^{-5}$, so $pK_a = - ext{log}(1.76 \times 10^{-5}) = 4.75$

Calculating pH:

$ext{pH} = 4.75 + \text{log} \left( \frac{0.015}{0.360} \right) = 4.28$