The salt solution that is basic in nature is: 

1. Ammonium chloride.

2. Ammonium sulphate.

3. Ammonium nitrate.

4. Sodium acetate.

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 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?