A hydrogen bromide solution, HBr(aq), reacts with water according to the following balanced chemical equation: HBr(aq) + H2O(l) ⇌ Br⁻(aq) + H3O⁺(aq) The Ka value of HBr(aq) at 25 °C is 1 × 10⁹ - NSC Physical Sciences - Question 7 - 2019 - Paper 2

The two bases in the reaction are:

1. Br⁻ (bromide ion)
2. H2O (water)

To find the pH of the solution after neutralization, first calculate the moles of NaOH used:

$n(NaOH) = c imes V = 0.5 imes 0.0165 = 0.00825 ext{ mol}$

Since NaOH is a strong base, it completely reacts with the leftover HBr. The moles of HBr initially was:

$n(HBr) = c imes V = 0.45 imes 0.09 = 0.0405 ext{ mol}$

The moles of HBr that reacted with NaOH is equal to the moles of NaOH, thus:

$n(HBr)_{excess} = n(HBr)_{initial} - n(NaOH) = 0.0405 - 0.00825 = 0.03225 ext{ mol}$

Now, we find the concentration of the leftover H⁺ ions in the total volume of the solution (90 cm³ + 16.5 cm³ = 106.5 cm³ or 0.1065 dm³):

c(H^+) = rac{n(HBr)_{excess}}{V_{total}} = rac{0.03225}{0.1065} ext{ mol·dm}^{-3} ext{ which is approximately } 0.303 ext{ mol·dm}^{-3}

Lastly, calculate the pH:

$pH = - ext{log}[H^+] = - ext{log}(0.303) ext{ which is approximately } 0.518$

Using the stoichiometry from the reaction:

$Zn(OH)_2 + 2HBr → 2Br^- + 2H_2O$

The ratio indicates that 1 mole of Zn(OH)₂ reacts with 2 moles of HBr. First, calculate the initial moles of HBr:

$n(HBr) = 0.45 imes 0.09 = 0.0405 ext{ mol}$

Thus, the moles of Zn(OH)₂ initially present:

n(Zn(OH)_2) = rac{1}{2} n(HBr) = rac{1}{2} imes 0.0405 = 0.02025 ext{ mol}

Finally, calculate the mass of Zn(OH)₂ using its molar mass (approximately 99 g/mol):

$ext{mass} = n imes ext{molar mass} = 0.02025 ext{ mol} imes 99 ext{ g/mol} = 2.0025 ext{ g}$