(a) (i) Write an expression for the solubility product, Ksp, of Ag2S(s) - CIE - A-Level Chemistry - Question 5 - 2019 - Paper 1

Given that the solubility of Ag2S(s) is 1.16 × 10−17 mol dm−3, we can substitute this into the Ksp expression:

Let $[Ag^+] = 2x$ and $[S^{2-}] = x$, where $x = 1.16 imes 10^{-17}$ mol dm^{-3}.

Therefore:

$K_{sp} = (2x)^2(x) = 4x^3$

Now, substituting the value of $x$:

$K_{sp} = 4(1.16 imes 10^{-17})^3 = 4 imes 1.575936 imes 10^{-51} = 6.303744 imes 10^{-51}$

Hence, $K_{sp} = 6.30 imes 10^{-51}$.

To find the volume of water needed to dissolve 1.00 g of Ag2S(s), we first need to calculate the moles of Ag2S:

$ext{Molar mass of Ag2S} = 2(107.87) + 32.07 = 247.81 ext{ g/mol}$

$ext{Moles of Ag2S} = \frac{1.00 ext{ g}}{247.81 ext{ g/mol}} = 0.00403 ext{ mol}$

From the solubility of Ag2S in water, we know that 1.16 × 10−17 mol/dm³ is the concentration. Now, we apply the formula:

$ext{Volume} = \frac{ ext{Moles}}{ ext{Concentration}} = \frac{0.00403}{1.16 imes 10^{-17}} ext{ dm}^3$

Calculating this gives:

$ext{Volume} = 3.47 imes 10^{15} ext{ dm}^3$

Given that HOBr is a weak acid with a dissociation constant of $K_a = 2.0 imes 10^{-4}$ mol dm−3,

Using the formula:

$K_a = \frac{[H^+][OBr^-]}{[HOBR]}$

Substituting the initial concentration:

Since $[H^+] = [OBr^-] = x$ and $[HOBR] = 0.20 - x$, we can approximate it as:

$K_a = \frac{x^2}{0.20}$

$2.0 imes 10^{-4} = \frac{x^2}{0.20}$

Solving for $x$ gives:

$x^2 = 4.0 imes 10^{-5}$ $x = 6.32 imes 10^{-3} ext{ mol/dm}^3$

The pH can then be calculated as:

$pH = -\log(6.32 imes 10^{-3}) = 2.20$

After adding KOH to the HOBr solution:

Moles of KOH = $0.20 ext{ mol/dm}^3 imes 0.005 ext{ dm}^3 = 0.001$ mol

Moles of HOBr remained = $0.20 ext{ mol/dm}^3 imes 0.020 ext{ dm}^3 - 0.001 = 0.003$ mol

Using the Henderson-Hasselbalch equation:

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

Where $pK_a = -\log(2.0 imes 10^{-4}) = 3.70$. The concentration of the weak acid ($[HA]$) and its conjugate base ($[A^-]$) in the final solution:

The total volume now is 20.0 + 5.0 = 25.0 cm³ (0.025 dm³).

$[HOBr] = \frac{0.003}{0.025} = 0.12 ext{ mol/dm}^3$ $[OBr^-] = \frac{0.001}{0.025} = 0.04 ext{ mol/dm}^3$

Now substituting in the Henderson-Hasselbalch formula:

$pH = 3.70 + \log \left( \frac{0.04}{0.12} \right) = 3.70 + ( -0.40) = 3.30$