The solubility product for a salt of type AB is $4\times {10}^{-8}$.  The molarity of its standard solution will be:
1. $2\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

2. $16\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

3. $2\times {10}^{-16}$ $\mathrm{mol}/\mathrm{L}$

4. $4\times {10}^{-4}$ $\mathrm{mol}/\mathrm{L}$

Given that the equilibrium constant for the reaction 

$2S{O}_{2\left(g\right)}$ $+$ $O_{2\left(g\right)}$ $\leftrightharpoons$ $2S{O}_{3\left(g\right)}$ $$

has a value of 278 at a particular temperature, the value of the equilibrium constant for the following reaction at the same temperature will be:

$S{O}_{3\left(g\right)}$ $$ $\leftrightharpoons$ $S{O}_{2\left(g\right)}$ $+$ $1/2$ $O_{2\left(g\right)}$ $$

1.  $3.6$ $\times$ ${10}^{-3}$ $$

2.  $6.0$ $\times$ ${10}^{-2}$ $$

3.  $1.3$ $\times$ ${10}^{-5}$ $$

4.  $1.8$ $\times$ ${10}^{-3}$ $$

Consider the following reaction:
A₂(g) + B₂(g)  ⇋ 2AB(g)  
At equilibrium, the concentrations of  A₂ = 3.0×10^–3 M;  B₂ = 4.2×10^–3 M and AB = 2.8×10^–3M.
The value \(K_C\)  for the above-given reaction in a sealed container at 527°C is:

In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag⁺ and Pb²⁺ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl^– concentration is 0.10 M. What will the concentration of Ag⁺ and Pb²⁺ at equilibrium?

(K_sp for AgCl  = 1.8 × 10^-10)$$
(K_sp for PbCl₂ = 1.7 × 10^-5) 

The reaction-

$2A+B\left(g\right)$ $\rightleftharpoons$ $3C\left(g\right)+D\left(g\right)$

begins with the concentrations of A and B both at an initial value of 1.00 M. When equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression:

1. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(0.50)}^2\right(0.75\left)\right]$

2. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(0.50)}^2\right(0.25\left)\right]$

3. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(0.75)}^2\right(0.25\left)\right]$

4. $\left[{(0.75)}^3\right(0.25\left)\right]÷\left[{(1.00)}^2\right(1.00\left)\right]$

The salt solution that is basic in nature is: 

1. Ammonium chloride.

2. Ammonium sulphate.

3. Ammonium nitrate.

4. Sodium acetate.

The maximum concentration of equimolar solutions, of ferrous sulphate and sodium sulphide, so that when mixed in equal volumes, there is no precipitation of iron sulphide, will be:

(For iron sulphide, K_sp = 6.3 × 10^–18). 

$1.$ $5.02$ $\times {10}^{-9}$ $\mathrm{M}$

$2.$ $5.02$ $\times$ ${10}^9$ $\mathrm{M}$

$3.$ $2.$ $25$ $\times$ ${10}^{-13}$ $\mathrm{M}$

$4.$ $\mathrm{Can}\text{'}\mathrm{t}$ $\mathrm{predict}$

The ionic product of water at 310 K is 2.7 × 10^–14.

The pH of neutral water at this temperature will be:

What will be the pH of a 0.1 M chloroacetic acid solution with an ionization constant of 1.35 × 10^–3?