The solubility product constant K${}_{\mathrm{sp}}$ of Mg(OH)${}_2$ is 9.0 × 10${}^{-12}$. If a solution is 0.010 M with respect to Mg${}^{2+}$ ion, what is the maximum hydroxide ion concentration which could be present without causing the precipitation of Mg(OH)${}_2$

1. 1.5 x 10${}^{-7}$ M

2. 3.0 x 10${}^{-7}$ M

3. 1.5 x 10${}^{-5}$ M

4. 3.0 x 10${}^{-5}$ M

For the equilibrium:

$LiCl·3N{H}_3\rightleftharpoons LiCl·N{H}_3+2N{H}_3·K_P=9<mspace\; width="1em"></mspace>at{m}^2$

At $27°C,$ a 8.2L vessel contains 0.1 mol of LiCl.$N{H}_3\left(s\right)$. How many moles of $N{H}_3$ should be added to the flask at this temperature of derive the backward reaction for completion?

(1) 20.5 mol

(2) 3 mol

(3) 1.2 mol

(4) 0.21 mol

Ammonia at a pressure 10 atm and $C{O}_2$ at a pressure at 20 atm are introduced into an evacuated chamber. If $K_p$ for the reaction

$N{H}_{2}COON{H}_4\left(s\right)<mspace\; width="1em"></mspace>\rightleftharpoons 2N{H}_3\left(g\right)+C{O}_2\left(g\right)$ is 2020 $at{m}^2$ at 400K then the final pressure in the chamber would be

(1) Less than 30 atm

(2) More than 30 atm

(3) Equal to 30 atm

(4) Unpredictable

A weak base BOH with $K_{b<mspace\; width="1em"></mspace>}={10}^{-5}$ is titrated with a strong acid HCl. At $\frac{3}{4}th$ of the equivalence point pH of solution is

(1) 5+log3

(2) 5-log3

(3) 9.425

(4) 8.523

The exact concentration of $H^{+}$ ion in ${10}^{-3}$ M aq. HCl solution at $25°C$ is:

1. ${10}^{-3}$

2. ${10}^{-3}+{10}^{-7}$

3. ${10}^{-3}+\frac{Kw}{\left[H^{+}\right]}$

4. ${10}^{-3}+\frac{Kw}{\left[O{H}^{-}\right]}$

A 0.1 molar solution of weak base BOH is 1% dissociated. If 0.2 moles of BCl is added in 1 litre solution of BOH, the degree of dissociation of BOH will become

1. $2\times {10}^{-3}$

2. $5\times {10}^{-5}$

3. 0.05

4. 0.02

Assuming no change in volume, calculate the minimum mass of NaCl (approx.) necessary to dissolve 0.01 mol AgCl in 100 L solution.

[Given $K_{sp}\left(AgCl\right)={10}^{-10};<mspace\; width="1em"></mspace>K_f\left(AgC{l}_2^{-}\right)=3\times {10}^5]$

1. 15 kg

2. 21 kg

3. 10 kg

4. 19kg

Henderson’s equation is pH = pK${}_{\mathrm{a}}$ + log$\frac{\left[\mathrm{salt}\right]}{\left[\mathrm{acid}\right]}$. If
the acid gets half neutralized the value of pH will
be : (pK${}_{\mathrm{a}}$= 4.30)

1. 4.3

2. 2.15

3. 8.60

4. 7

For ${\mathrm{N}}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>3{\mathrm{H}}_2<mspace\; width="1em"></mspace>\rightleftharpoons <mspace\; width="1em"></mspace>2{\mathrm{NH}}_3$ , 1 mol ${\mathrm{N}}_2$ and 3 mol ${\mathrm{H}}_2$ are at 4 atm. Equilibrium pressure is found to be 3 atm. Hence, ${\mathrm{K}}_{\mathrm{p}}$
is

1. $\frac{1}{(0.5){(0.15)}^3}$

2. $\frac{1}{(0.5){(1.5)}^3}$

3. $\frac{3<mspace\; width="1em"></mspace>\mathrm{x}<mspace\; width="1em"></mspace>3}{(0.5){(1.5)}^3}$

4. None of these

PCl${}_5$(g) $\rightleftharpoons$ PCl${}_3$(g) + Cl${}_2$(g) if the degree of the
dissociation is $\mathrm{\alpha }$ at equilibrium pressure P, then
the equilibrium constant for the reaction

1. ${\mathrm{K}}_{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{\alpha }}^2}{1+<mspace\; width="1em"></mspace>{\mathrm{\alpha }}^2\mathrm{P}}$

2. ${\mathrm{K}}_{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{\alpha }}^2{\mathrm{P}}^2}{1-{\mathrm{\alpha }}^2}$

3. ${\mathrm{K}}_{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{\alpha P}}^2}{1-{\mathrm{\alpha }}^2}$

4. ${\mathrm{K}}_{\mathrm{p}}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>\frac{{\mathrm{\alpha }}^2\mathrm{P}}{1-{\mathrm{\alpha }}^2}$