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

First, we need to calculate the moles of HCl and Ba(OH)₂:

• Moles of HCl:

  0.100 , \text{mol dm}^{-3} \times \frac{10.35 , \text{cm}^3}{1000} = 0.001035 , \text{mol}

• Moles of Ba(OH)₂:

  0.150 , \text{mol dm}^{-3} \times \frac{25.0 , \text{cm}^3}{1000} = 0.00375 , \text{mol}

Each mole of Ba(OH)₂ provides 2 moles of OH⁻ ions:

0.00375 , \text{mol} \times 2 = 0.00750 , \text{mol} \text{ of } OH^-\n Now, determine the total volume after mixing 10.35 cm³ of HCl and 25.0 cm³ of Ba(OH)₂:

10.35 , \text{cm}^3 + 25.0 , \text{cm}^3 = 35.35 , \text{cm}^3 = 0.03535 , \text{dm}^3

Now, calculate the concentration of OH⁻:

[OH^-] = \frac{0.00750 , \text{mol}}{0.03535 , \text{dm}^3} = 0.2124 , \text{mol dm}^{-3}

Calculate the pOH:

pOH = -\log(0.2124) = 0.6708\n
Calculate the pH:

pH + pOH = 14
pH = 14 - 0.6708 = 13.3292\

Therefore, the pH of the solution is approximately 13.33.

Water is neutral at this temperature because the concentrations of hydrogen ions [H⁺] and hydroxide ions [OH⁻] are equal, resulting in a pH of 7.

SO₂ is the oxide that reacts with water to form sulfurous acid, leading to an acidic solution.

The expression for the acid dissociation constant for ethanoic acid is given by:

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

1. Calculate the initial moles of sodium ethanoate:

0.025 , \text{mol} \text{ in } 500 , \text{cm}^3 = 0.00005 , \text{mol dm}^{-3}

2. Calculate the moles of ethanoic acid:

0.700 , \text{mol dm}^{-3} \times 0.500 , \text{dm}^3 = 0.350 , \text{mol}

3. Calculate the initial moles of HCl added:

2.00 , \text{mol dm}^{-3} \times \frac{5.00 , \text{cm}^3}{1000} = 0.0100 , \text{mol}

4. Update the moles of the weak acid (ethanoic acid):

0.350 , \text{mol} - 0.0100 , \text{mol} = 0.340 , \text{mol}

5. Use the Henderson-Hasselbalch equation to find the pH:

pH = pK_a + \log \frac{[A^-]}{[HA]} = \log \frac{0.025}{0.340} + 4.76 d