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.
Similar Questions
- X and Y are the number of electrons involved, respectively during the oxidation of $\text{I}^-$ to $\text{I}_2$ and $\text{S}^{2-}$ to S by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. The value of X + Y is ______.
- Electricity is passed through an acidic solution of $\text{Cu}^{2+}$ till all the $\text{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)
[Given:
$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \quad \text{E}^\circ_{\text{red}} = +0.34 \text{ V}$
$\text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \text{E}^\circ_{\text{red}} = +1.23 \text{ V}$
Molar mass of Cu = $63.54 \text{ g mol}^{-1}$
Molar mass of $\text{O}_2$ = $32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = 22.4 L] - The half-life of ${}^{65}\text{Zn}$ is 245 days. After x days, 75% of original activity remained. The value of $x$ in days is _______. (Nearest integer)
(Given: $\log 3 = 0.4771$ and $\log 2 = 0.3010$) - Molar conductivity of a weak acid HQ of concentration $0.18 \text{ M}$ was found to be $1/30$ of the molar conductivity of another weak acid HZ with concentration of $0.02 \text{ M}$. If $\lambda^\circ_{Q^-}$ happened to be equal with $\lambda^\circ_{Z^-}$, then the difference of the $\text{pK}_a$ values of the two weak acids ($\text{pK}_a\text{(HQ)} - \text{pK}_a\text{(HZ)}$) is _______ (Nearest integer).
[Given: degree of dissociation ($\alpha$) < < 1 for both weak acids, $\lambda^\circ$ : limiting molar conductivity of ions] - The first and second ionization constants of $\text{H}_2\text{X}$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $\text{X}^{2-}$ in $0.1\text{ M}$ $\text{H}_2\text{X}$ solution is __________ $\times 10^{-15}\text{ M}$. (Nearest Integer)
- The osmotic pressure of a living cell is $12\text{ atm}$ at $300\text{ K}$. The strength of sodium chloride solution that is isotonic with the living cell at this temperature is __________ $\text{g L}^{-1}$. (Nearest integer)
Given : $\text{R} = 0.08\text{ L atm K}^{-1}\text{ mol}^{-1}$
Assume complete dissociation of $\text{NaCl}$
(Given : Molar mass of $\text{Na}$ and $\text{Cl}$ are $23$ and $35.5\text{ g mol}^{-1}$ respectively.) - $MX$ is a sparingly soluble salt that follows the given solubility equilibrium at $298\text{ K}$.
$\text{MX (s)} \rightleftharpoons \text{M}^+\text{ (aq)} + \text{X}^-\text{ (aq)}; \quad \text{K}_{\text{sp}} = 10^{-10}$
If the standard reduction potential for $\text{M}^+\text{(aq)} \xrightarrow{+\text{e}^-} \text{M(s)}$ is $\left(\text{E}^\ominus_{\text{M}^+/\text{M}}\right) = 0.79\text{ V}$, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $\text{E}^\ominus_{\text{X}^-/\text{MX (s)}/\text{M}}$ is __________ $\text{mV}$. (nearest integer)
[Given : $\frac{2.303\text{ RT}}{\text{F}} = 0.059\text{ V}$] - For strong electrolyte $\Lambda_{\text{m}}$ increases slowly with dilution and can be represented by the equation $\Lambda_{\text{m}} = \Lambda_{\text{m}}^{\circ} - \text{A}\text{c}^{1/2}$
Molar conductivity values of the solutions of strong electrolyte AB at $18^{\circ}\text{C}$ are given below :$\text{c } [\text{mol L}^{-1}]$ $0.04$ $0.09$ $0.16$ $0.25$ $\Lambda_{\text{m}} \ [\text{S cm}^{2}\text{ mol}^{-1}]$ $96.1$ $95.7$ $95.3$ $94.9$
The value of constant A based on the above data $[\text{in S cm}^{2}\text{ mol}^{-1} / (\text{mol}/\text{L})^{1/2}]$ unit is _________. - $\text{A} \longrightarrow \text{B} \text{ (first reaction)}$
$\text{C} \longrightarrow \text{D} \text{ (second reaction)}$
Consider the above two first-order reactions. The rate constant for first reaction at $500\text{ K}$ is double of the same at $300\text{ K}$. At $500\text{ K}$, $50\%$ of the reaction becomes complete in $2\text{ hour}$. The activation energy of the second reaction is half of that of first reaction. If the rate constant at $500\text{ K}$ of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at $300\text{ K}$ is _________ $\times 10^{-1}\text{hour}^{-1}$ (nearest integer). For the thermal decomposition of reactant $\text{AB(g)}$, the following plot is constructed.
The half life of the reaction is '$x$' min.
$x = $ ___________ min. (Nearest integer)
Important Questions from Physical Chemistry
- X and Y are the number of electrons involved, respectively during the oxidation of $\text{I}^-$ to $\text{I}_2$ and $\text{S}^{2-}$ to S by acidified $\text{K}_2\text{Cr}_2\text{O}_7$. The value of X + Y is ______.
- Electricity is passed through an acidic solution of $\text{Cu}^{2+}$ till all the $\text{Cu}^{2+}$ was exhausted, leading to the deposition of 300 mg of Cu metal. However, a current of 600 mA was continued to pass through the same solution for another 28 minutes by keeping the total volume of the solution fixed at 200 mL. The total volume of oxygen evolved at STP during the entire process is ______ mL. (Nearest integer)
[Given:
$\text{Cu}^{2+}\text{(aq)} + 2\text{e}^- \rightarrow \text{Cu(s)} \quad \text{E}^\circ_{\text{red}} = +0.34 \text{ V}$
$\text{O}_2\text{(g)} + 4\text{H}^+ + 4\text{e}^- \rightarrow 2\text{H}_2\text{O} \quad \text{E}^\circ_{\text{red}} = +1.23 \text{ V}$
Molar mass of Cu = $63.54 \text{ g mol}^{-1}$
Molar mass of $\text{O}_2$ = $32 \text{ g mol}^{-1}$
Faraday Constant = $96500 \text{ C mol}^{-1}$
Molar volume at STP = 22.4 L] - The half-life of ${}^{65}\text{Zn}$ is 245 days. After x days, 75% of original activity remained. The value of $x$ in days is _______. (Nearest integer)
(Given: $\log 3 = 0.4771$ and $\log 2 = 0.3010$) - Molar conductivity of a weak acid HQ of concentration $0.18 \text{ M}$ was found to be $1/30$ of the molar conductivity of another weak acid HZ with concentration of $0.02 \text{ M}$. If $\lambda^\circ_{Q^-}$ happened to be equal with $\lambda^\circ_{Z^-}$, then the difference of the $\text{pK}_a$ values of the two weak acids ($\text{pK}_a\text{(HQ)} - \text{pK}_a\text{(HZ)}$) is _______ (Nearest integer).
[Given: degree of dissociation ($\alpha$) < < 1 for both weak acids, $\lambda^\circ$ : limiting molar conductivity of ions] - The first and second ionization constants of $\text{H}_2\text{X}$ are $2.5 \times 10^{-8}$ and $1.0 \times 10^{-13}$ respectively. The concentration of $\text{X}^{2-}$ in $0.1\text{ M}$ $\text{H}_2\text{X}$ solution is __________ $\times 10^{-15}\text{ M}$. (Nearest Integer)
