The salt that gives a neutral solution in water is: 

1.	KBr	2.	NH4NO3
3.	NaCN	4.	Rb2(CO3)

The ionization constant of benzoic acid is 6.46×10⁵ and K_sp for silver benzoate is 2.5×10^–13. Silver benzoate is x times more soluble in a buffer of pH 3.19 compared to its solubility in pure water. The value of x will be:

1. 6.8 

2. 16.8 

3. 33.3 

4.  3.3 

Consider the following reaction:
\(\mathrm{Cu}\left(\mathrm{NO}_3\right)_2(\mathrm{~s}) \rightleftharpoons 2 \mathrm{CuO}(\mathrm{~s})+4 \mathrm{NO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g})\)

What is the equilibrium constant \(K_c\)​ expression for the above reaction?

$1. {\mathrm{K}}_{\mathrm{C}}= \frac{{\left[{\mathrm{NO}}_2\right]}^4\left[{\mathrm{O}}_2\right]}{\left[\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2\right]}$
$2. {\mathrm{K}}_{\mathrm{C}}= {\left[{\mathrm{NO}}_2\right]}^4\left[{\mathrm{O}}_2\right]$
$3. {\mathrm{K}}_{\mathrm{C}} =\frac{{\left[\mathrm{CuO}\right]}^2{\left[{\mathrm{NO}}_2\right]}^4\left[{\mathrm{O}}_2\right]}{\left[\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2\right]}$
$4. {\mathrm{K}}_{\mathrm{C}} =\frac{{\left[{\mathrm{NO}}_2\right]}^4\left[{\mathrm{O}}_2\right]}{\left[\mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2\right]}$

${{\mathrm{I}}_2}_{\left(\mathrm{s}\right)}$ $+$ $5{{\mathrm{F}}_2}_{\left(\mathrm{g}\right)}$ $$ $\to 2{\mathrm{IF}}_{5\left(g\right)}$ $$

The solution that has the lowest pH is: 

(assuming 100% dissociation)

The degree of ionization of a 0.1 M bromoacetic acid solution is 0.132. The pK_a of bromacetic acid will be: 

1. 1.98

2. 2.75

3. 4.56

4. 7.15

The species that can act as Bronsted acids as well as bases is/are:

1.  ${\mathrm{H}}_2\mathrm{O}$

2.  ${\mathrm{HCO}}_3^{-},$ ${\mathrm{HSO}}_4^{-}$ $$

3.  ${\mathrm{NH}}_3$

4.  All of the above.

The solubility product of silver chromate is $1.1\times {10}^{-12}$. The solubility of silver chromate will be:

1. $6.5\times {10}^{-5}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

2. $6.5\times {10}^{-6}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

3. $5.5\times {10}^{-5}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

4. $5.5\times {10}^{-6}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

The solubility product of mercurous iodide is $4.5\times {10}^{-29}$. The solubility of mercurous iodide will be:

1. $6.5\times {10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

2. $4.09\times {10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

3. $4.09\times {10}^{-7}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

4. $6.5\times {10}^{-8}\mathrm{mol}$ ${\mathrm{L}}^{-1}$

Consider the following graph:

The point that represents reaction in equilibrium is-

1. C

2. A

3. B

4. D