Consider two Group IV metal ions $\text{X}^{2+}$ and $\text{Y}^{2+}$.A solution containing 0.01 M $\text{X}^{2+}$ and 0.01 M $\text{Y}^{2+}$ is saturated with $\text{H}_2\text{S}$. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer)(Given: $\text{K}_{sp}(\text{XS}) = 1 \times 10^{-22}$ at $25^\circ\text{C}$, $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$ at $25^\circ\text{C}$, $[\text{H}_2\text{S}] = 0.1\text{M}$ in solution, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$)

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Consider two Group IV metal ions $\text{X}^{2+}$ and $\text{Y}^{2+}$.
A solution containing 0.01 M $\text{X}^{2+}$ and 0.01 M $\text{Y}^{2+}$ is saturated with $\text{H}_2\text{S}$. The pH at which the metal sulphide YS will form as a precipitate is ______. (Nearest integer)
(Given: $\text{K}_{sp}(\text{XS}) = 1 \times 10^{-22}$ at $25^\circ\text{C}$, $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$ at $25^\circ\text{C}$, $[\text{H}_2\text{S}] = 0.1\text{M}$ in solution, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$)

Precipitation Condition for YS

Precipitation of the metal sulfide YS begins when the ion product, $[\text{Y}^{2+}][\text{S}^{2-}]$, becomes equal to or greater than its solubility product constant, $\text{K}_{sp}(\text{YS})$. We calculate the condition for the onset of precipitation:

$[\text{Y}^{2+}][\text{S}^{2-}] = \text{K}_{sp}(\text{YS})$

Required Sulfide Ion Concentration

Given the initial concentration of $\text{Y}^{2+}$ ions is $0.01 \text{ M}$ and $\text{K}_{sp}(\text{YS}) = 4 \times 10^{-16}$.

The minimum concentration of sulfide ions, $[\text{S}^{2-}]$, required for YS to start precipitating is:

$[\text{S}^{2-}] = \frac{\text{K}_{sp}(\text{YS})}{[\text{Y}^{2+}]} = \frac{4 \times 10^{-16}}{0.01 \text{ M}} = \frac{4 \times 10^{-16}}{1 \times 10^{-2}} = 4 \times 10^{-14} \text{ M}$

Relating Sulfide Concentration to pH

The concentration of sulfide ions $[\text{S}^{2-}]$ in a solution saturated with $\text{H}_2\text{S}$ is dependent on the $\text{pH}$. The overall dissociation of hydrogen sulfide is:

$\text{H}_2\text{S} \rightleftharpoons 2\text{H}^+ + \text{S}^{2-}$

The equilibrium constant for this reaction is $K = K_{a1} \times K_{a2} = 1.0 \times 10^{-21}$. The relationship is expressed as:

$K = \frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]}$

We can rearrange this formula to solve for the hydrogen ion concentration, $[\text{H}^+]$:

$[\text{H}^+]^2 = \frac{K \times [\text{H}_2\text{S}]}{[\text{S}^{2-}]}$

Calculating Hydrogen Ion Concentration

Substitute the given values into the rearranged equation:

Equilibrium constant, $K = 1.0 \times 10^{-21}$
Concentration of $\text{H}_2\text{S}$, $[\text{H}_2\text{S}] = 0.1 \text{ M} = 1 \times 10^{-1} \text{ M}$
Required sulfide ion concentration, $[\text{S}^{2-}] = 4 \times 10^{-14} \text{ M}$

$[\text{H}^+]^2 = \frac{(1.0 \times 10^{-21}) \times (1 \times 10^{-1})}{4 \times 10^{-14}}$

$[\text{H}^+]^2 = \frac{1.0 \times 10^{-22}}{4 \times 10^{-14}} = 0.25 \times 10^{-8} = 2.5 \times 10^{-9}$

Determining the pH

Calculate the hydrogen ion concentration $[\text{H}^+]$ by taking the square root:

$[\text{H}^+] = \sqrt{2.5 \times 10^{-9}} = \sqrt{25 \times 10^{-10}} = 5 \times 10^{-5} \text{ M}$

Now, calculate the pH using the definition $\text{pH} = -\log[\text{H}^+]$:

$\text{pH} = -\log(5 \times 10^{-5})$

Using $\log 5 = 0.70$:

$\text{pH} = -(\log 5 + \log 10^{-5}) = -(0.70 - 5) = 5 - 0.70 = 4.30$

The question asks for the $\text{pH}$ value rounded to the nearest integer. The calculated $\text{pH}$ is 4.30.

The nearest integer to 4.30 is 4.

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E. Primary standard should have low relative molar mass.
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Consider $\text{A} \xrightarrow{k_1} \text{B}$ and $\text{C} \xrightarrow{k_2} \text{D}$ are two reactions. If the rate constant ($k_1$) of the $\text{A} \longrightarrow \text{B}$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T/K}$ and activation energy of $\text{C} \longrightarrow \text{D}$ reaction ($\text{Ea}_2$) is $\frac{1}{5}$th of the $\text{A} \longrightarrow \text{B}$ reaction ($\text{Ea}_1$), then the value of ($\text{Ea}_2$) is _________ $\text{kJ mol}^{-1}$. (Nearest Integer)
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