229839
The $\mathrm{pH}$ of a solution containing $0.10 \mathrm{M}$ sodium acetate and $0.03 \mathrm{M}$ acetic acid is $\left(\mathrm{pK}_{\mathrm{n}}\right.$ for $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right)$
1 4.09
2 6.09
3 5.09
4 7.09
Explanation:
According to Henderson's equation $\begin{aligned} \mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log \left(\frac{\text { salt }}{\text { acid }}\right) \\ \mathrm{CH}_3 \mathrm{COOH} & +\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \end{aligned}$ Given value, $\mathrm{pK}_{\mathrm{a}}=4.57$ Sodium acetate concentration $=0.10 \mathrm{M}$ acetic acid concentration $=0.03 \mathrm{M}$ Putting the value we get $\mathrm{pH}=4.57+\log \frac{0.10}{0.03}$ $\mathrm{pH}=5.09$
AP-EAMCET-(2000)
Ionic Equilibrium
229840
Calculate the $\mathrm{pOH}$ of $0.10 \mathrm{M} \mathrm{HCl}$ solution.
229841
One litre of an aqueous solution has $3.65 \mathrm{~g}$ of $\mathrm{HCl}$. If it is desired to increase the $\mathrm{pH}$ of the solution to 2 then $\mathrm{H}_3 \mathrm{O}^{+}$ion concentration present initially should be
1 also doubled
2 reduced to half
3 increased by 10 times
4 reduced by 10 times.
Explanation:
Given - $\text { amount of aqueous solution }=1 \mathrm{~L}$ $\text { amount of } \mathrm{HCl} \quad=3.65 \mathrm{~g}$ $\text { Molar mass of } \mathrm{HCl}=36.5$ $\therefore \text { Molarity }=\frac{3.65}{36.5 \times 1}=\frac{1}{10} \mathrm{M}$ $\mathrm{pH}=-\log \left[\frac{1}{10}\right]=1$ $\text { Given }-\mathrm{pH}=2$ $=-\log _{10}\left[\mathrm{H}^{+}\right]=2$ $\frac{1}{[\mathrm{H}]^{+}}=100$ $\text { or } \quad\left[\mathrm{H}^{+}\right]=\frac{1}{10 \Omega} \mathrm{M}$ From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.
229839
The $\mathrm{pH}$ of a solution containing $0.10 \mathrm{M}$ sodium acetate and $0.03 \mathrm{M}$ acetic acid is $\left(\mathrm{pK}_{\mathrm{n}}\right.$ for $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right)$
1 4.09
2 6.09
3 5.09
4 7.09
Explanation:
According to Henderson's equation $\begin{aligned} \mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log \left(\frac{\text { salt }}{\text { acid }}\right) \\ \mathrm{CH}_3 \mathrm{COOH} & +\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \end{aligned}$ Given value, $\mathrm{pK}_{\mathrm{a}}=4.57$ Sodium acetate concentration $=0.10 \mathrm{M}$ acetic acid concentration $=0.03 \mathrm{M}$ Putting the value we get $\mathrm{pH}=4.57+\log \frac{0.10}{0.03}$ $\mathrm{pH}=5.09$
AP-EAMCET-(2000)
Ionic Equilibrium
229840
Calculate the $\mathrm{pOH}$ of $0.10 \mathrm{M} \mathrm{HCl}$ solution.
229841
One litre of an aqueous solution has $3.65 \mathrm{~g}$ of $\mathrm{HCl}$. If it is desired to increase the $\mathrm{pH}$ of the solution to 2 then $\mathrm{H}_3 \mathrm{O}^{+}$ion concentration present initially should be
1 also doubled
2 reduced to half
3 increased by 10 times
4 reduced by 10 times.
Explanation:
Given - $\text { amount of aqueous solution }=1 \mathrm{~L}$ $\text { amount of } \mathrm{HCl} \quad=3.65 \mathrm{~g}$ $\text { Molar mass of } \mathrm{HCl}=36.5$ $\therefore \text { Molarity }=\frac{3.65}{36.5 \times 1}=\frac{1}{10} \mathrm{M}$ $\mathrm{pH}=-\log \left[\frac{1}{10}\right]=1$ $\text { Given }-\mathrm{pH}=2$ $=-\log _{10}\left[\mathrm{H}^{+}\right]=2$ $\frac{1}{[\mathrm{H}]^{+}}=100$ $\text { or } \quad\left[\mathrm{H}^{+}\right]=\frac{1}{10 \Omega} \mathrm{M}$ From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.
229839
The $\mathrm{pH}$ of a solution containing $0.10 \mathrm{M}$ sodium acetate and $0.03 \mathrm{M}$ acetic acid is $\left(\mathrm{pK}_{\mathrm{n}}\right.$ for $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right)$
1 4.09
2 6.09
3 5.09
4 7.09
Explanation:
According to Henderson's equation $\begin{aligned} \mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log \left(\frac{\text { salt }}{\text { acid }}\right) \\ \mathrm{CH}_3 \mathrm{COOH} & +\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \end{aligned}$ Given value, $\mathrm{pK}_{\mathrm{a}}=4.57$ Sodium acetate concentration $=0.10 \mathrm{M}$ acetic acid concentration $=0.03 \mathrm{M}$ Putting the value we get $\mathrm{pH}=4.57+\log \frac{0.10}{0.03}$ $\mathrm{pH}=5.09$
AP-EAMCET-(2000)
Ionic Equilibrium
229840
Calculate the $\mathrm{pOH}$ of $0.10 \mathrm{M} \mathrm{HCl}$ solution.
229841
One litre of an aqueous solution has $3.65 \mathrm{~g}$ of $\mathrm{HCl}$. If it is desired to increase the $\mathrm{pH}$ of the solution to 2 then $\mathrm{H}_3 \mathrm{O}^{+}$ion concentration present initially should be
1 also doubled
2 reduced to half
3 increased by 10 times
4 reduced by 10 times.
Explanation:
Given - $\text { amount of aqueous solution }=1 \mathrm{~L}$ $\text { amount of } \mathrm{HCl} \quad=3.65 \mathrm{~g}$ $\text { Molar mass of } \mathrm{HCl}=36.5$ $\therefore \text { Molarity }=\frac{3.65}{36.5 \times 1}=\frac{1}{10} \mathrm{M}$ $\mathrm{pH}=-\log \left[\frac{1}{10}\right]=1$ $\text { Given }-\mathrm{pH}=2$ $=-\log _{10}\left[\mathrm{H}^{+}\right]=2$ $\frac{1}{[\mathrm{H}]^{+}}=100$ $\text { or } \quad\left[\mathrm{H}^{+}\right]=\frac{1}{10 \Omega} \mathrm{M}$ From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.
229839
The $\mathrm{pH}$ of a solution containing $0.10 \mathrm{M}$ sodium acetate and $0.03 \mathrm{M}$ acetic acid is $\left(\mathrm{pK}_{\mathrm{n}}\right.$ for $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right)$
1 4.09
2 6.09
3 5.09
4 7.09
Explanation:
According to Henderson's equation $\begin{aligned} \mathrm{pH} & =\mathrm{pK}_{\mathrm{a}}+\log \left(\frac{\text { salt }}{\text { acid }}\right) \\ \mathrm{CH}_3 \mathrm{COOH} & +\mathrm{NaOH} \rightarrow \mathrm{CH}_3 \mathrm{COONa}+\mathrm{H}_2 \mathrm{O} \end{aligned}$ Given value, $\mathrm{pK}_{\mathrm{a}}=4.57$ Sodium acetate concentration $=0.10 \mathrm{M}$ acetic acid concentration $=0.03 \mathrm{M}$ Putting the value we get $\mathrm{pH}=4.57+\log \frac{0.10}{0.03}$ $\mathrm{pH}=5.09$
AP-EAMCET-(2000)
Ionic Equilibrium
229840
Calculate the $\mathrm{pOH}$ of $0.10 \mathrm{M} \mathrm{HCl}$ solution.
229841
One litre of an aqueous solution has $3.65 \mathrm{~g}$ of $\mathrm{HCl}$. If it is desired to increase the $\mathrm{pH}$ of the solution to 2 then $\mathrm{H}_3 \mathrm{O}^{+}$ion concentration present initially should be
1 also doubled
2 reduced to half
3 increased by 10 times
4 reduced by 10 times.
Explanation:
Given - $\text { amount of aqueous solution }=1 \mathrm{~L}$ $\text { amount of } \mathrm{HCl} \quad=3.65 \mathrm{~g}$ $\text { Molar mass of } \mathrm{HCl}=36.5$ $\therefore \text { Molarity }=\frac{3.65}{36.5 \times 1}=\frac{1}{10} \mathrm{M}$ $\mathrm{pH}=-\log \left[\frac{1}{10}\right]=1$ $\text { Given }-\mathrm{pH}=2$ $=-\log _{10}\left[\mathrm{H}^{+}\right]=2$ $\frac{1}{[\mathrm{H}]^{+}}=100$ $\text { or } \quad\left[\mathrm{H}^{+}\right]=\frac{1}{10 \Omega} \mathrm{M}$ From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.
