The ionization constant of \(CH_{3} COOH\) is 1.7 × \(\left(10\right)^{- 5}\) and the concentration of \(H^{+ }\)ions is 3.4 × \(\left(10\right)^{- 4}\). The initial concentration of \(CH_{3} COOH\) will be:
1.  3.4 × \(\left(10\right)^{- 4}\)
2.  3.4× \(\left(10\right)^{- 3}\)
3.  6.8 × \(\left(10\right)^{- 4}\)
4.  6.8 × \(\left(10\right)^{- 3}\)

Find the solubility product (Kₛₚ) of M₂S having a molar solubility of 3.5 × 10⁻⁶ mol L⁻¹.

1. 1.7×10^-6 mol³ litre^-3

2. 1.7×10^-16 mol³ litre^-3

3. 1.7×10^-18 mol³ litre^-3

4. 1.7×10^-12 mol³ litre^-3

The pK_b of dimethylamine and pk_a of acetic acid are 3.27 and 4.77 respectively at T (K).
The correct option for the pH of dimethylammonium acetate solution is:

The following equilibria are given:

The equilibrium constant of the reaction
\(2NH_{3} \ + \ \frac{5}{2}O_{2} \ \rightleftharpoons \ 2NO \ + \ 3H_{2}O\) in terms of K₁, K₂ and K₃ is:

1. K₁.K₂.K₃
2. \(\mathrm{\frac{K_{1}K_{2}}{K_{3}}}\)
3. \(\mathrm{\frac{K_{1}K_{3}^{2}}{K_{3}}}\)
4. \(\mathrm{\frac{K_{2}K_{3}^{3}}{K_{1}}}\)

Calculate the solubility of AgI in a 10⁻⁴ N KI solution at 25°C, given that the solubility product constant (Kₛₚ) of AgI is 1.0 × 10⁻¹⁶ mol² L⁻².

1. $1.0\times {10}^{-16}$

2. $1.0\times {10}^{-12}$

3. $1.0\times {10}^{-10}$

4. $1.0\times {10}^{-8}$

The reaction quotient (Q) for the reaction:

${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$

is given by $\mathrm{Q}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}$.
Under what conditions will the reaction proceed from right to left?

1. Q = K_C
2. Q < K_C
3. Q > K_C
4. Q = 0

(where K_C is the equilibrium constant)

A compound, among the following, that cannot be classified as a protonic acid is:

If a solution of 0.1 N NH₄OH and 0.1 N NH₄Cl has pH 9.25, then pK_b of NH₄OH is:

1. 9.25

2. 4.75

3. 3.75

4. 8.25

The compound with the highest pH among the following is:

1. CH₃COOK

2. Na₂CO₃

3. NH₄Cl

4. NaNO₃

If the solubility of MX₂ – type electrolytes is 0.5 × 10^–4 Mole/lit. then K_sp of electrolytes will be:

1. $5\times {10}^{-12}$

2. $25\times {10}^{-10}$

3. $1\times {10}^{-13}$

4. $5\times {10}^{-13}$