229808
The dissociation constant of a weak base is $1 \times 10^{-5}$ at $25^{\circ} \mathrm{C}$. The $\mathrm{pH}$ of its $0.1 \mathrm{M}$ solution at the same temperature will be
229810
The $\mathrm{pOH}$ of $0.0005 \mathrm{M}$ sulphuric acid is
1 5
2 3
3 11
4 12
Explanation:
The equation for ionization is- $\mathrm{H}_2 \mathrm{SO}_4 \square \quad 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}$ Each molecule of acid gives 2 molecules of $\mathrm{H}^{+}$ So, $\left[\mathrm{H}^{+}\right]=2 \times 0.0005=1 \times 10^{-3} \mathrm{M}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-3}\right)=3$ We know that- $\begin{aligned} & \because \mathrm{pH}+\mathrm{pOH}=14 \\ & 3+\mathrm{pOH}=14 \\ & \mathrm{pOH}=14-3 \\ & \mathrm{pOH}=11 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229811
The $\mathrm{pH}$ of $\mathrm{HCl}$ is 5 . If $10 \mathrm{~mL}$ of this solution is diluted to $100 \mathrm{~mL}$, the $\mathrm{pH}$ of the resultant solution is
1 5.1
2 6.9
3 11
4 12
Explanation:
Given that, $\mathrm{pH}$ of $\mathrm{HCl}=5$ So, Concentration of $\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{~mol} / \mathrm{L}$ Concentration of $\mathrm{H}^{+}$ions in $10 \mathrm{ml}$ of the solution $\begin{aligned} & =\frac{10^{-5} \times 10}{1000} \\ & =10^{-7} \mathrm{~mol} \end{aligned}$ We know that, solution is diluted to $100 \mathrm{ml}$ i.e. 10 times No. of $\mathrm{H}^{+}$ions $=\frac{10^{-7}}{10}=10^{-8}$ Total $\left[\mathrm{H}^{+}\right]=10^{-8}+10^{-7}$ $\left(\left[\mathrm{H}^{+}\right]\right.$from water cannot be neglected.) $\begin{aligned} & =10^{-8}(1+10)=11 \times 10^{-8} \mathrm{M} \\ & \mathrm{pH}=-\log \left(11 \times 10^{-8}\right)=-(1.0413-8)=6.95 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229812
The $\mathrm{pH}$ of a mixture of $10 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4, 5$ $\mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{HCl}$ and $5 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2$ is
1 1
2 0.5
3 0
4 1.5
Explanation:
{l} |Exp: Molarity of mixture of $10 _2 _4=0.1 $ | |---| |$=1$ millimole of $_2 _4$ | |$=2$ millimole of $^{-}$ions | |$5 $ of $0.2 =1$ millimole of $^{+}$ions | |$5 $ of $0.1 ()_2=0.5$ millimole of $()_2$ | |$=1$ millimole of $^{-}$ions | |1 millimole of $^{-}$ions will neutralize the 1 millimole | |of $^{+}$ions. | |So, 2 millimoles of $^{+}$ions will remain in solution. | |{$[^{+}] {20}=0.1=10^{-1}$ mol/ } | |Hence, $=- [^{+}]=- (10^{-1})=1$
229808
The dissociation constant of a weak base is $1 \times 10^{-5}$ at $25^{\circ} \mathrm{C}$. The $\mathrm{pH}$ of its $0.1 \mathrm{M}$ solution at the same temperature will be
229810
The $\mathrm{pOH}$ of $0.0005 \mathrm{M}$ sulphuric acid is
1 5
2 3
3 11
4 12
Explanation:
The equation for ionization is- $\mathrm{H}_2 \mathrm{SO}_4 \square \quad 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}$ Each molecule of acid gives 2 molecules of $\mathrm{H}^{+}$ So, $\left[\mathrm{H}^{+}\right]=2 \times 0.0005=1 \times 10^{-3} \mathrm{M}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-3}\right)=3$ We know that- $\begin{aligned} & \because \mathrm{pH}+\mathrm{pOH}=14 \\ & 3+\mathrm{pOH}=14 \\ & \mathrm{pOH}=14-3 \\ & \mathrm{pOH}=11 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229811
The $\mathrm{pH}$ of $\mathrm{HCl}$ is 5 . If $10 \mathrm{~mL}$ of this solution is diluted to $100 \mathrm{~mL}$, the $\mathrm{pH}$ of the resultant solution is
1 5.1
2 6.9
3 11
4 12
Explanation:
Given that, $\mathrm{pH}$ of $\mathrm{HCl}=5$ So, Concentration of $\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{~mol} / \mathrm{L}$ Concentration of $\mathrm{H}^{+}$ions in $10 \mathrm{ml}$ of the solution $\begin{aligned} & =\frac{10^{-5} \times 10}{1000} \\ & =10^{-7} \mathrm{~mol} \end{aligned}$ We know that, solution is diluted to $100 \mathrm{ml}$ i.e. 10 times No. of $\mathrm{H}^{+}$ions $=\frac{10^{-7}}{10}=10^{-8}$ Total $\left[\mathrm{H}^{+}\right]=10^{-8}+10^{-7}$ $\left(\left[\mathrm{H}^{+}\right]\right.$from water cannot be neglected.) $\begin{aligned} & =10^{-8}(1+10)=11 \times 10^{-8} \mathrm{M} \\ & \mathrm{pH}=-\log \left(11 \times 10^{-8}\right)=-(1.0413-8)=6.95 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229812
The $\mathrm{pH}$ of a mixture of $10 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4, 5$ $\mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{HCl}$ and $5 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2$ is
1 1
2 0.5
3 0
4 1.5
Explanation:
{l} |Exp: Molarity of mixture of $10 _2 _4=0.1 $ | |---| |$=1$ millimole of $_2 _4$ | |$=2$ millimole of $^{-}$ions | |$5 $ of $0.2 =1$ millimole of $^{+}$ions | |$5 $ of $0.1 ()_2=0.5$ millimole of $()_2$ | |$=1$ millimole of $^{-}$ions | |1 millimole of $^{-}$ions will neutralize the 1 millimole | |of $^{+}$ions. | |So, 2 millimoles of $^{+}$ions will remain in solution. | |{$[^{+}] {20}=0.1=10^{-1}$ mol/ } | |Hence, $=- [^{+}]=- (10^{-1})=1$
229808
The dissociation constant of a weak base is $1 \times 10^{-5}$ at $25^{\circ} \mathrm{C}$. The $\mathrm{pH}$ of its $0.1 \mathrm{M}$ solution at the same temperature will be
229810
The $\mathrm{pOH}$ of $0.0005 \mathrm{M}$ sulphuric acid is
1 5
2 3
3 11
4 12
Explanation:
The equation for ionization is- $\mathrm{H}_2 \mathrm{SO}_4 \square \quad 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}$ Each molecule of acid gives 2 molecules of $\mathrm{H}^{+}$ So, $\left[\mathrm{H}^{+}\right]=2 \times 0.0005=1 \times 10^{-3} \mathrm{M}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-3}\right)=3$ We know that- $\begin{aligned} & \because \mathrm{pH}+\mathrm{pOH}=14 \\ & 3+\mathrm{pOH}=14 \\ & \mathrm{pOH}=14-3 \\ & \mathrm{pOH}=11 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229811
The $\mathrm{pH}$ of $\mathrm{HCl}$ is 5 . If $10 \mathrm{~mL}$ of this solution is diluted to $100 \mathrm{~mL}$, the $\mathrm{pH}$ of the resultant solution is
1 5.1
2 6.9
3 11
4 12
Explanation:
Given that, $\mathrm{pH}$ of $\mathrm{HCl}=5$ So, Concentration of $\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{~mol} / \mathrm{L}$ Concentration of $\mathrm{H}^{+}$ions in $10 \mathrm{ml}$ of the solution $\begin{aligned} & =\frac{10^{-5} \times 10}{1000} \\ & =10^{-7} \mathrm{~mol} \end{aligned}$ We know that, solution is diluted to $100 \mathrm{ml}$ i.e. 10 times No. of $\mathrm{H}^{+}$ions $=\frac{10^{-7}}{10}=10^{-8}$ Total $\left[\mathrm{H}^{+}\right]=10^{-8}+10^{-7}$ $\left(\left[\mathrm{H}^{+}\right]\right.$from water cannot be neglected.) $\begin{aligned} & =10^{-8}(1+10)=11 \times 10^{-8} \mathrm{M} \\ & \mathrm{pH}=-\log \left(11 \times 10^{-8}\right)=-(1.0413-8)=6.95 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229812
The $\mathrm{pH}$ of a mixture of $10 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4, 5$ $\mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{HCl}$ and $5 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2$ is
1 1
2 0.5
3 0
4 1.5
Explanation:
{l} |Exp: Molarity of mixture of $10 _2 _4=0.1 $ | |---| |$=1$ millimole of $_2 _4$ | |$=2$ millimole of $^{-}$ions | |$5 $ of $0.2 =1$ millimole of $^{+}$ions | |$5 $ of $0.1 ()_2=0.5$ millimole of $()_2$ | |$=1$ millimole of $^{-}$ions | |1 millimole of $^{-}$ions will neutralize the 1 millimole | |of $^{+}$ions. | |So, 2 millimoles of $^{+}$ions will remain in solution. | |{$[^{+}] {20}=0.1=10^{-1}$ mol/ } | |Hence, $=- [^{+}]=- (10^{-1})=1$
229808
The dissociation constant of a weak base is $1 \times 10^{-5}$ at $25^{\circ} \mathrm{C}$. The $\mathrm{pH}$ of its $0.1 \mathrm{M}$ solution at the same temperature will be
229810
The $\mathrm{pOH}$ of $0.0005 \mathrm{M}$ sulphuric acid is
1 5
2 3
3 11
4 12
Explanation:
The equation for ionization is- $\mathrm{H}_2 \mathrm{SO}_4 \square \quad 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}$ Each molecule of acid gives 2 molecules of $\mathrm{H}^{+}$ So, $\left[\mathrm{H}^{+}\right]=2 \times 0.0005=1 \times 10^{-3} \mathrm{M}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-3}\right)=3$ We know that- $\begin{aligned} & \because \mathrm{pH}+\mathrm{pOH}=14 \\ & 3+\mathrm{pOH}=14 \\ & \mathrm{pOH}=14-3 \\ & \mathrm{pOH}=11 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229811
The $\mathrm{pH}$ of $\mathrm{HCl}$ is 5 . If $10 \mathrm{~mL}$ of this solution is diluted to $100 \mathrm{~mL}$, the $\mathrm{pH}$ of the resultant solution is
1 5.1
2 6.9
3 11
4 12
Explanation:
Given that, $\mathrm{pH}$ of $\mathrm{HCl}=5$ So, Concentration of $\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{~mol} / \mathrm{L}$ Concentration of $\mathrm{H}^{+}$ions in $10 \mathrm{ml}$ of the solution $\begin{aligned} & =\frac{10^{-5} \times 10}{1000} \\ & =10^{-7} \mathrm{~mol} \end{aligned}$ We know that, solution is diluted to $100 \mathrm{ml}$ i.e. 10 times No. of $\mathrm{H}^{+}$ions $=\frac{10^{-7}}{10}=10^{-8}$ Total $\left[\mathrm{H}^{+}\right]=10^{-8}+10^{-7}$ $\left(\left[\mathrm{H}^{+}\right]\right.$from water cannot be neglected.) $\begin{aligned} & =10^{-8}(1+10)=11 \times 10^{-8} \mathrm{M} \\ & \mathrm{pH}=-\log \left(11 \times 10^{-8}\right)=-(1.0413-8)=6.95 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229812
The $\mathrm{pH}$ of a mixture of $10 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4, 5$ $\mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{HCl}$ and $5 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2$ is
1 1
2 0.5
3 0
4 1.5
Explanation:
{l} |Exp: Molarity of mixture of $10 _2 _4=0.1 $ | |---| |$=1$ millimole of $_2 _4$ | |$=2$ millimole of $^{-}$ions | |$5 $ of $0.2 =1$ millimole of $^{+}$ions | |$5 $ of $0.1 ()_2=0.5$ millimole of $()_2$ | |$=1$ millimole of $^{-}$ions | |1 millimole of $^{-}$ions will neutralize the 1 millimole | |of $^{+}$ions. | |So, 2 millimoles of $^{+}$ions will remain in solution. | |{$[^{+}] {20}=0.1=10^{-1}$ mol/ } | |Hence, $=- [^{+}]=- (10^{-1})=1$
229808
The dissociation constant of a weak base is $1 \times 10^{-5}$ at $25^{\circ} \mathrm{C}$. The $\mathrm{pH}$ of its $0.1 \mathrm{M}$ solution at the same temperature will be
229810
The $\mathrm{pOH}$ of $0.0005 \mathrm{M}$ sulphuric acid is
1 5
2 3
3 11
4 12
Explanation:
The equation for ionization is- $\mathrm{H}_2 \mathrm{SO}_4 \square \quad 2 \mathrm{H}^{+}+\mathrm{SO}_4^{2-}$ Each molecule of acid gives 2 molecules of $\mathrm{H}^{+}$ So, $\left[\mathrm{H}^{+}\right]=2 \times 0.0005=1 \times 10^{-3} \mathrm{M}$ $\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log \left(1 \times 10^{-3}\right)=3$ We know that- $\begin{aligned} & \because \mathrm{pH}+\mathrm{pOH}=14 \\ & 3+\mathrm{pOH}=14 \\ & \mathrm{pOH}=14-3 \\ & \mathrm{pOH}=11 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229811
The $\mathrm{pH}$ of $\mathrm{HCl}$ is 5 . If $10 \mathrm{~mL}$ of this solution is diluted to $100 \mathrm{~mL}$, the $\mathrm{pH}$ of the resultant solution is
1 5.1
2 6.9
3 11
4 12
Explanation:
Given that, $\mathrm{pH}$ of $\mathrm{HCl}=5$ So, Concentration of $\left[\mathrm{H}^{+}\right]=10^{-5} \mathrm{~mol} / \mathrm{L}$ Concentration of $\mathrm{H}^{+}$ions in $10 \mathrm{ml}$ of the solution $\begin{aligned} & =\frac{10^{-5} \times 10}{1000} \\ & =10^{-7} \mathrm{~mol} \end{aligned}$ We know that, solution is diluted to $100 \mathrm{ml}$ i.e. 10 times No. of $\mathrm{H}^{+}$ions $=\frac{10^{-7}}{10}=10^{-8}$ Total $\left[\mathrm{H}^{+}\right]=10^{-8}+10^{-7}$ $\left(\left[\mathrm{H}^{+}\right]\right.$from water cannot be neglected.) $\begin{aligned} & =10^{-8}(1+10)=11 \times 10^{-8} \mathrm{M} \\ & \mathrm{pH}=-\log \left(11 \times 10^{-8}\right)=-(1.0413-8)=6.95 \end{aligned}$
COMEDK-2011
Ionic Equilibrium
229812
The $\mathrm{pH}$ of a mixture of $10 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4, 5$ $\mathrm{ml}$ of $0.2 \mathrm{M} \mathrm{HCl}$ and $5 \mathrm{ml}$ of $0.1 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2$ is
1 1
2 0.5
3 0
4 1.5
Explanation:
{l} |Exp: Molarity of mixture of $10 _2 _4=0.1 $ | |---| |$=1$ millimole of $_2 _4$ | |$=2$ millimole of $^{-}$ions | |$5 $ of $0.2 =1$ millimole of $^{+}$ions | |$5 $ of $0.1 ()_2=0.5$ millimole of $()_2$ | |$=1$ millimole of $^{-}$ions | |1 millimole of $^{-}$ions will neutralize the 1 millimole | |of $^{+}$ions. | |So, 2 millimoles of $^{+}$ions will remain in solution. | |{$[^{+}] {20}=0.1=10^{-1}$ mol/ } | |Hence, $=- [^{+}]=- (10^{-1})=1$
