Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)
Degree of dissociation of $0.1\,N\,\,C{H_3}COOH$ is (Dissociation constant $ = 1 \times {10^{ - 5}}$)
- A
${10^{ - 5}}$
- B
${10^{ - 4}}$
- C
${10^{ - 3}}$
- D
${10^{ - 2}}$
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${K_{C{H_3}COOH}} = 1.9 \times {10^{ - 5}}$. Calculate $pH$ at end point in titration of $0.1$ $M$ $C{H_3}COOH$ and $0.1$ $M$ $NaOH$.
${K_{C{H_3}COOH}} = 1.9 \times {10^{ - 5}}$. Calculate $pH$ at end point in titration of $0.1$ $M$ $C{H_3}COOH$ and $0.1$ $M$ $NaOH$.
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[ ${H_2}S$ of ${K_a}\left( 1 \right) = 1.0 \times {10^{ - 7}}$ and ${K_a}\left( 2 \right) = 1.3 \times {10^{ - 13}}$ ]
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[ ${H_2}S$ of ${K_a}\left( 1 \right) = 1.0 \times {10^{ - 7}}$ and ${K_a}\left( 2 \right) = 1.3 \times {10^{ - 13}}$ ]
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