229147
The hydrogen ion concentration in $\mathrm{mol} / \mathrm{dm}^{3}$, in a $0.2 \mathrm{~m}$ solution of a weak acid, HA $\left(\mathrm{K}_{\mathrm{a}}=\mathbf{2} \times 10^{-5}\right)$ is close to
229148
The degree of dissociation of an acid $\mathrm{HA}$ in 0.1 $M$ solution is $0.1 \%$. Its dissociation constant is
1 $1 \times 10^{-3}$
2 $1 \times 10^{-7}$
3 $1 \times 10^{-10}$
4 $1 \times 10^{-14}$
Explanation:
Given that, degree of dissociation $(\alpha)=0.1 \%$ $=10^{-3}$ $\mathrm{C}=0.1 \mathrm{M}=10^{-1} \mathrm{M}$ According to ostwald dilution law, the dissociation constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ is equal to the product of molar concentration $(\mathrm{C})$ and the square of the degree of dissociation $\left(\alpha^{2}\right)$. $\begin{aligned} \therefore \text { Dissociation constant }(\mathrm{K}) & =\alpha^{2} . \mathrm{C} \\ & =\left(10^{-3}\right)^{2} \times 10^{-1} \\ & =10^{-7} \end{aligned}$
AP-EAMCET-1992
Chemical Equilibrium
229149
$100 \mathrm{~mL}$ of a solution contains $2 \mathrm{~g}$ of acetic acid and $3 \mathrm{~g}$ of sodium acetate providing $K_{a}=1.8 \times 10^{-5}$, then choose the correct option
1 The solution is basic in nature
2 The solution is acidic in nature
3 The solution is amphoteric in nature
4 The solution is neutral in nature JIMPER-2016
Explanation:
According to Henderson's equation- \begin{aligned} & \mathrm{pH}=-\log _{10} \mathrm{~K}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]} \\ & {[\text { Salt }]=\frac{3 \times 1000}{82 \times 100} \mathrm{molL}^{-1} \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COONa} \text { is salt and } \\ \text { molecular mass of this } \\ \text { salt is } 82 \end{array}\right]} \end{aligned} \([\text { Acid }]=\frac{2 \times 1000}{60 \times 100} \text { mol L }^1 \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COOH} \text { is acid and } \\ \text { molecular mass of this } \\ \text { acid is } 60 \end{array}\right]\) Given, \(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\) \(\begin{aligned} & \therefore \quad \mathrm{pH}=-\log _{10} 1.8 \times 10^{-5}+\log _{10}\left[\frac{\frac{3 \times 1000}{82 \times 100}}{\frac{2 \times 1000}{60 \times 100}}\right] \\ & \mathrm{pH}=4.7851 \end{aligned}\) The solution is a acidic nature.
Chemical Equilibrium
229151
The dissociation of water at $25^{\circ} \mathrm{C}$ is $1.9 \times 10^{-7} \%$ and the density of water is $1.0 \mathrm{~g} / \mathrm{cm}^{3}$ the ionization constant of water is
229147
The hydrogen ion concentration in $\mathrm{mol} / \mathrm{dm}^{3}$, in a $0.2 \mathrm{~m}$ solution of a weak acid, HA $\left(\mathrm{K}_{\mathrm{a}}=\mathbf{2} \times 10^{-5}\right)$ is close to
229148
The degree of dissociation of an acid $\mathrm{HA}$ in 0.1 $M$ solution is $0.1 \%$. Its dissociation constant is
1 $1 \times 10^{-3}$
2 $1 \times 10^{-7}$
3 $1 \times 10^{-10}$
4 $1 \times 10^{-14}$
Explanation:
Given that, degree of dissociation $(\alpha)=0.1 \%$ $=10^{-3}$ $\mathrm{C}=0.1 \mathrm{M}=10^{-1} \mathrm{M}$ According to ostwald dilution law, the dissociation constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ is equal to the product of molar concentration $(\mathrm{C})$ and the square of the degree of dissociation $\left(\alpha^{2}\right)$. $\begin{aligned} \therefore \text { Dissociation constant }(\mathrm{K}) & =\alpha^{2} . \mathrm{C} \\ & =\left(10^{-3}\right)^{2} \times 10^{-1} \\ & =10^{-7} \end{aligned}$
AP-EAMCET-1992
Chemical Equilibrium
229149
$100 \mathrm{~mL}$ of a solution contains $2 \mathrm{~g}$ of acetic acid and $3 \mathrm{~g}$ of sodium acetate providing $K_{a}=1.8 \times 10^{-5}$, then choose the correct option
1 The solution is basic in nature
2 The solution is acidic in nature
3 The solution is amphoteric in nature
4 The solution is neutral in nature JIMPER-2016
Explanation:
According to Henderson's equation- \begin{aligned} & \mathrm{pH}=-\log _{10} \mathrm{~K}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]} \\ & {[\text { Salt }]=\frac{3 \times 1000}{82 \times 100} \mathrm{molL}^{-1} \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COONa} \text { is salt and } \\ \text { molecular mass of this } \\ \text { salt is } 82 \end{array}\right]} \end{aligned} \([\text { Acid }]=\frac{2 \times 1000}{60 \times 100} \text { mol L }^1 \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COOH} \text { is acid and } \\ \text { molecular mass of this } \\ \text { acid is } 60 \end{array}\right]\) Given, \(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\) \(\begin{aligned} & \therefore \quad \mathrm{pH}=-\log _{10} 1.8 \times 10^{-5}+\log _{10}\left[\frac{\frac{3 \times 1000}{82 \times 100}}{\frac{2 \times 1000}{60 \times 100}}\right] \\ & \mathrm{pH}=4.7851 \end{aligned}\) The solution is a acidic nature.
Chemical Equilibrium
229151
The dissociation of water at $25^{\circ} \mathrm{C}$ is $1.9 \times 10^{-7} \%$ and the density of water is $1.0 \mathrm{~g} / \mathrm{cm}^{3}$ the ionization constant of water is
229147
The hydrogen ion concentration in $\mathrm{mol} / \mathrm{dm}^{3}$, in a $0.2 \mathrm{~m}$ solution of a weak acid, HA $\left(\mathrm{K}_{\mathrm{a}}=\mathbf{2} \times 10^{-5}\right)$ is close to
229148
The degree of dissociation of an acid $\mathrm{HA}$ in 0.1 $M$ solution is $0.1 \%$. Its dissociation constant is
1 $1 \times 10^{-3}$
2 $1 \times 10^{-7}$
3 $1 \times 10^{-10}$
4 $1 \times 10^{-14}$
Explanation:
Given that, degree of dissociation $(\alpha)=0.1 \%$ $=10^{-3}$ $\mathrm{C}=0.1 \mathrm{M}=10^{-1} \mathrm{M}$ According to ostwald dilution law, the dissociation constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ is equal to the product of molar concentration $(\mathrm{C})$ and the square of the degree of dissociation $\left(\alpha^{2}\right)$. $\begin{aligned} \therefore \text { Dissociation constant }(\mathrm{K}) & =\alpha^{2} . \mathrm{C} \\ & =\left(10^{-3}\right)^{2} \times 10^{-1} \\ & =10^{-7} \end{aligned}$
AP-EAMCET-1992
Chemical Equilibrium
229149
$100 \mathrm{~mL}$ of a solution contains $2 \mathrm{~g}$ of acetic acid and $3 \mathrm{~g}$ of sodium acetate providing $K_{a}=1.8 \times 10^{-5}$, then choose the correct option
1 The solution is basic in nature
2 The solution is acidic in nature
3 The solution is amphoteric in nature
4 The solution is neutral in nature JIMPER-2016
Explanation:
According to Henderson's equation- \begin{aligned} & \mathrm{pH}=-\log _{10} \mathrm{~K}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]} \\ & {[\text { Salt }]=\frac{3 \times 1000}{82 \times 100} \mathrm{molL}^{-1} \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COONa} \text { is salt and } \\ \text { molecular mass of this } \\ \text { salt is } 82 \end{array}\right]} \end{aligned} \([\text { Acid }]=\frac{2 \times 1000}{60 \times 100} \text { mol L }^1 \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COOH} \text { is acid and } \\ \text { molecular mass of this } \\ \text { acid is } 60 \end{array}\right]\) Given, \(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\) \(\begin{aligned} & \therefore \quad \mathrm{pH}=-\log _{10} 1.8 \times 10^{-5}+\log _{10}\left[\frac{\frac{3 \times 1000}{82 \times 100}}{\frac{2 \times 1000}{60 \times 100}}\right] \\ & \mathrm{pH}=4.7851 \end{aligned}\) The solution is a acidic nature.
Chemical Equilibrium
229151
The dissociation of water at $25^{\circ} \mathrm{C}$ is $1.9 \times 10^{-7} \%$ and the density of water is $1.0 \mathrm{~g} / \mathrm{cm}^{3}$ the ionization constant of water is
229147
The hydrogen ion concentration in $\mathrm{mol} / \mathrm{dm}^{3}$, in a $0.2 \mathrm{~m}$ solution of a weak acid, HA $\left(\mathrm{K}_{\mathrm{a}}=\mathbf{2} \times 10^{-5}\right)$ is close to
229148
The degree of dissociation of an acid $\mathrm{HA}$ in 0.1 $M$ solution is $0.1 \%$. Its dissociation constant is
1 $1 \times 10^{-3}$
2 $1 \times 10^{-7}$
3 $1 \times 10^{-10}$
4 $1 \times 10^{-14}$
Explanation:
Given that, degree of dissociation $(\alpha)=0.1 \%$ $=10^{-3}$ $\mathrm{C}=0.1 \mathrm{M}=10^{-1} \mathrm{M}$ According to ostwald dilution law, the dissociation constant $\left(\mathrm{K}_{\mathrm{a}}\right)$ is equal to the product of molar concentration $(\mathrm{C})$ and the square of the degree of dissociation $\left(\alpha^{2}\right)$. $\begin{aligned} \therefore \text { Dissociation constant }(\mathrm{K}) & =\alpha^{2} . \mathrm{C} \\ & =\left(10^{-3}\right)^{2} \times 10^{-1} \\ & =10^{-7} \end{aligned}$
AP-EAMCET-1992
Chemical Equilibrium
229149
$100 \mathrm{~mL}$ of a solution contains $2 \mathrm{~g}$ of acetic acid and $3 \mathrm{~g}$ of sodium acetate providing $K_{a}=1.8 \times 10^{-5}$, then choose the correct option
1 The solution is basic in nature
2 The solution is acidic in nature
3 The solution is amphoteric in nature
4 The solution is neutral in nature JIMPER-2016
Explanation:
According to Henderson's equation- \begin{aligned} & \mathrm{pH}=-\log _{10} \mathrm{~K}_{\mathrm{a}}+\log _{10} \frac{[\text { Salt }]}{[\text { Base }]} \\ & {[\text { Salt }]=\frac{3 \times 1000}{82 \times 100} \mathrm{molL}^{-1} \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COONa} \text { is salt and } \\ \text { molecular mass of this } \\ \text { salt is } 82 \end{array}\right]} \end{aligned} \([\text { Acid }]=\frac{2 \times 1000}{60 \times 100} \text { mol L }^1 \quad\left[\begin{array}{l} \mathrm{CH}_3 \mathrm{COOH} \text { is acid and } \\ \text { molecular mass of this } \\ \text { acid is } 60 \end{array}\right]\) Given, \(\mathrm{K}_{\mathrm{a}}=1.8 \times 10^{-5}\) \(\begin{aligned} & \therefore \quad \mathrm{pH}=-\log _{10} 1.8 \times 10^{-5}+\log _{10}\left[\frac{\frac{3 \times 1000}{82 \times 100}}{\frac{2 \times 1000}{60 \times 100}}\right] \\ & \mathrm{pH}=4.7851 \end{aligned}\) The solution is a acidic nature.
Chemical Equilibrium
229151
The dissociation of water at $25^{\circ} \mathrm{C}$ is $1.9 \times 10^{-7} \%$ and the density of water is $1.0 \mathrm{~g} / \mathrm{cm}^{3}$ the ionization constant of water is
