$$BiO(OH) (s) \rightleftharpoons BiO^+ (aq) + OH^- (aq)$$
$K = 4 \times 10^{-10}$
(Given : $\log 2 = 0.3010$)
Equilibrium Calculation for $BiO(OH)$ Dissolution
The dissolution of $BiO(OH)(s)$ establishes the following equilibrium:
$BiO(OH) (s) \rightleftharpoons BiO^+ (aq) + OH^- (aq)$The equilibrium constant ($K$) expression is given by:
$K = [BiO^+][OH^-]$We are given $K = 4 \times 10^{-10}$.
Determining Hydroxide Ion Concentration
Since the solid $BiO(OH)$ does not appear in the equilibrium expression, and assuming initial concentrations of $BiO^+$ and $OH^-$ are zero before dissolution:
- Let $[OH^-] = x$.
- From the stoichiometry, $[BiO^+] = [OH^-] = x$.
- Substituting into the $K$ expression: $K = x \times x = x^2$.
Therefore:
$x^2 = 4 \times 10^{-10}$ $x = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5} \, \text{M}$So, the hydroxide ion concentration is $[OH^-] = 2 \times 10^{-5}$ M.
Calculating pH from pOH
First, calculate the pOH:
$\text{pOH} = -\log_{10}[OH^-]$ $\text{pOH} = -\log_{10}(2 \times 10^{-5})$ $\text{pOH} = -(\log_{10} 2 + \log_{10} 10^{-5})$ $\text{pOH} = -(\log_{10} 2 - 5)$Using the given value $\log_{10} 2 = 0.3010$:
$\text{pOH} = -(0.3010 - 5)$ $\text{pOH} = 5 - 0.3010$ $\text{pOH} = 4.699$At 298 K, the relationship between pH and pOH is:
$\text{pH} + \text{pOH} = 14$Now, calculate the pH:
$\text{pH} = 14 - \text{pOH}$ $\text{pH} = 14 - 4.699$ $\text{pH} = 9.301$Similar Questions
- A solution of copper sulphate is electrolysed for $10 \text{ minutes}$ with a current of $1.5 \text{ amperes}$. The mass of copper deposited at cathode is :
(Given : Molar mass of Cu = $63 \text{ g mol}^{-1}$, $1 \text{ F} = 96487 \text{ C mol}^{-1}$) - The number of hydrogen atoms present in $5.4 \text{ g}$ of urea is :
(Given : Molar mass of urea : $60 \text{ g mol}^{-1}$, $N_A : 6.022 \times 10^{23} \text{ particles mol}^{-1}$) - Match List I with List II :
List I (Order of reaction) List II (Unit of rate constant) A. Zero order I. $mol^{-1} L s^{-1}$ B. First order II. $mol^{-2} L^2 s^{-1}$ C. Second order III. $s^{-1}$ D. Third order IV. $mol L^{-1} s^{-1}$
Choose the correct answer from the options given below : Consider the following reaction :
$2A (g) + B (g) \rightarrow 2D (g)$
$\Delta U^\ominus = -10 \text{ kJ mol}^{-1}$ and $\Delta S^\ominus = -44 \text{ J K}^{-1}$ at 298 K.
Identify the correct option with $\Delta G^\ominus$ for the reaction and spontaneity of the reaction at 298 K. (Given : $R = 8.31 \text{ J mol}^{-1} \text{K}^{-1}$)- A bulb is rated at 150 watt, converting 8% energy into light. If energy of one photon is $4.42 \times 10^{-19} \text{ J}$, how many photons are emitted by the bulb per second ?
For a certain reaction R $\rightarrow$ Product, the plot of [R] vs time has a negative slope as shown. The order of reaction is :
At 298 K, a certain buffer solution contains equal concentrations of $X^{-}$ and $HX$. $K_b$ for $X^-$ is $10^{-10}$. What is the pH of this buffer solution ?
- At a certain temperature T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is :
- Given below are certain reactions. Identify the reaction for which $K_p \neq K_c$.
- Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, T (K).
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k \text{ in s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)
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 ______.
- 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, $\text{K}_{a1} \times \text{K}_{a2}(\text{H}_2\text{S}) = 1.0 \times 10^{-21}$, $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 5 = 0.70$) - 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]
