What is the molarity of the standard solution if the solubility product for a salt of type AB is $4\times {10}^{-8}$?

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

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?

The ionization constant of nitrous acid is 4.5 × 10^–4.  The pH of a 0.04 M sodium nitrite solution will be:

The pH of a 0.1 M solution of cyanic acid (HCNO) is 2.34. The ionization constant of the acid will be:

$1.$ $2.02$ $\times$ ${10}^4$
$2.$ $3.14$ $\times$ ${10}^3$
$3.$ $2.02$ $\times$ ${10}^{-4}$
$4.$ $1.01$ $\times$ ${10}^{-4}$

0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. The pH of the solution will be