314996
HCOOH and \(\mathrm{CH}_{3} \mathrm{COOH}\) solutions have equal pH . If \(\mathrm{K}_{1} / \mathrm{K}_{2}\) is 4 , their molar concentration ratio will be
1 2
2 0.5
3 4
4 0.25
Explanation:
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{H}^{+}\right]=\mathrm{C} \alpha=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{C}}\) Given pH of two acids as same, so \(\sqrt{\mathrm{K}_{\mathrm{a}_{1}} \mathrm{C}_{1}}=\sqrt{\mathrm{K}_{\mathrm{a}_{2}} \mathrm{C}_{2}}\) \(\dfrac{\mathrm{K}_{\mathrm{a}_{1}}}{\mathrm{~K}_{\mathrm{a}_{2}}}=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) \(4=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) or \(\dfrac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\dfrac{1}{4}=0.25\)
CHXI07:EQUILIBRIUM
314997
When \({\text{50 mL 0}}{\text{.6 M HCl}}\) and \({\text{50 mL 0}}{\text{.3 M NaOH}}\) are mixed, the species with maximum concentration in the resulting solution is
314998
Equal volumes of \(0.1 \mathrm{M}\) potassium hydroxide and \(0.1 \mathrm{M}\) sulphuric acid are mixed. The \(\mathrm{pH}\) of resulting solution is
1 7
2 0
3 Less than 7
4 Greater than 7
Explanation:
m.eq. of acid \(\mathrm {=0.1 \times 2 \times \mathrm{V}}\) m.eq. of base \(\mathrm {=0.1 \times \mathrm{V}}\) Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.
CHXI07:EQUILIBRIUM
314999
\(5 \mathrm{~mL}\) of \({\rm{0}}{\rm{.4\;N}}\,{\rm{NaOH}}\) is mxied with \(20 \mathrm{~mL}\) of \({\rm{0}}{\rm{.1\;N}}\,\,{\rm{HCl}}\) The \(\mathrm{pH}\) of the resulting soluiton will be
1 6
2 7
3 8
4 5
Explanation:
Millimoles of \(\mathrm {\mathrm{NaOH}=5 \times 0.4=2}\) Millimoles of \(\mathrm {\mathrm{HCl}=20 \times 0.1=2}\) So, 2 millimoles of \(\mathrm {\mathrm{NaOH}}\) and 2 millimoles of \(\mathrm {\mathrm{HCl}}\) completely neutralise each other, resulting into a neutral solution \(\mathrm {(\mathrm{pH}=7)}\).
CHXI07:EQUILIBRIUM
315000
Determine the \(\mathrm{pH}\) of the solution that results from the addition of \(20.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) to \(30.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{HCl}\).
1 11.30
2 10.55
3 2.70
4 83.35
Explanation:
Millimoles of \(\mathrm{H}^{+}=30 \times 0.01=0.3\) Millimole of \(\mathrm{OH}^{-}=20 \times 0.01 \times 2=0.4\) Remaining millimoles of \(\mathrm{OH}^{-}\) \(\begin{aligned}= & 0.4-0.3=0.1 \\\Rightarrow\left[\mathrm{OH}^{-}\right] & =\dfrac{0.1}{50} \text { or } 2 \times 10^{-3}\end{aligned}\) So, \(\mathrm{pOH}=2.7\) \(\mathrm{pH}=14-2.7=11.30\)
314996
HCOOH and \(\mathrm{CH}_{3} \mathrm{COOH}\) solutions have equal pH . If \(\mathrm{K}_{1} / \mathrm{K}_{2}\) is 4 , their molar concentration ratio will be
1 2
2 0.5
3 4
4 0.25
Explanation:
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{H}^{+}\right]=\mathrm{C} \alpha=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{C}}\) Given pH of two acids as same, so \(\sqrt{\mathrm{K}_{\mathrm{a}_{1}} \mathrm{C}_{1}}=\sqrt{\mathrm{K}_{\mathrm{a}_{2}} \mathrm{C}_{2}}\) \(\dfrac{\mathrm{K}_{\mathrm{a}_{1}}}{\mathrm{~K}_{\mathrm{a}_{2}}}=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) \(4=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) or \(\dfrac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\dfrac{1}{4}=0.25\)
CHXI07:EQUILIBRIUM
314997
When \({\text{50 mL 0}}{\text{.6 M HCl}}\) and \({\text{50 mL 0}}{\text{.3 M NaOH}}\) are mixed, the species with maximum concentration in the resulting solution is
314998
Equal volumes of \(0.1 \mathrm{M}\) potassium hydroxide and \(0.1 \mathrm{M}\) sulphuric acid are mixed. The \(\mathrm{pH}\) of resulting solution is
1 7
2 0
3 Less than 7
4 Greater than 7
Explanation:
m.eq. of acid \(\mathrm {=0.1 \times 2 \times \mathrm{V}}\) m.eq. of base \(\mathrm {=0.1 \times \mathrm{V}}\) Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.
CHXI07:EQUILIBRIUM
314999
\(5 \mathrm{~mL}\) of \({\rm{0}}{\rm{.4\;N}}\,{\rm{NaOH}}\) is mxied with \(20 \mathrm{~mL}\) of \({\rm{0}}{\rm{.1\;N}}\,\,{\rm{HCl}}\) The \(\mathrm{pH}\) of the resulting soluiton will be
1 6
2 7
3 8
4 5
Explanation:
Millimoles of \(\mathrm {\mathrm{NaOH}=5 \times 0.4=2}\) Millimoles of \(\mathrm {\mathrm{HCl}=20 \times 0.1=2}\) So, 2 millimoles of \(\mathrm {\mathrm{NaOH}}\) and 2 millimoles of \(\mathrm {\mathrm{HCl}}\) completely neutralise each other, resulting into a neutral solution \(\mathrm {(\mathrm{pH}=7)}\).
CHXI07:EQUILIBRIUM
315000
Determine the \(\mathrm{pH}\) of the solution that results from the addition of \(20.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) to \(30.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{HCl}\).
1 11.30
2 10.55
3 2.70
4 83.35
Explanation:
Millimoles of \(\mathrm{H}^{+}=30 \times 0.01=0.3\) Millimole of \(\mathrm{OH}^{-}=20 \times 0.01 \times 2=0.4\) Remaining millimoles of \(\mathrm{OH}^{-}\) \(\begin{aligned}= & 0.4-0.3=0.1 \\\Rightarrow\left[\mathrm{OH}^{-}\right] & =\dfrac{0.1}{50} \text { or } 2 \times 10^{-3}\end{aligned}\) So, \(\mathrm{pOH}=2.7\) \(\mathrm{pH}=14-2.7=11.30\)
314996
HCOOH and \(\mathrm{CH}_{3} \mathrm{COOH}\) solutions have equal pH . If \(\mathrm{K}_{1} / \mathrm{K}_{2}\) is 4 , their molar concentration ratio will be
1 2
2 0.5
3 4
4 0.25
Explanation:
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{H}^{+}\right]=\mathrm{C} \alpha=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{C}}\) Given pH of two acids as same, so \(\sqrt{\mathrm{K}_{\mathrm{a}_{1}} \mathrm{C}_{1}}=\sqrt{\mathrm{K}_{\mathrm{a}_{2}} \mathrm{C}_{2}}\) \(\dfrac{\mathrm{K}_{\mathrm{a}_{1}}}{\mathrm{~K}_{\mathrm{a}_{2}}}=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) \(4=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) or \(\dfrac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\dfrac{1}{4}=0.25\)
CHXI07:EQUILIBRIUM
314997
When \({\text{50 mL 0}}{\text{.6 M HCl}}\) and \({\text{50 mL 0}}{\text{.3 M NaOH}}\) are mixed, the species with maximum concentration in the resulting solution is
314998
Equal volumes of \(0.1 \mathrm{M}\) potassium hydroxide and \(0.1 \mathrm{M}\) sulphuric acid are mixed. The \(\mathrm{pH}\) of resulting solution is
1 7
2 0
3 Less than 7
4 Greater than 7
Explanation:
m.eq. of acid \(\mathrm {=0.1 \times 2 \times \mathrm{V}}\) m.eq. of base \(\mathrm {=0.1 \times \mathrm{V}}\) Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.
CHXI07:EQUILIBRIUM
314999
\(5 \mathrm{~mL}\) of \({\rm{0}}{\rm{.4\;N}}\,{\rm{NaOH}}\) is mxied with \(20 \mathrm{~mL}\) of \({\rm{0}}{\rm{.1\;N}}\,\,{\rm{HCl}}\) The \(\mathrm{pH}\) of the resulting soluiton will be
1 6
2 7
3 8
4 5
Explanation:
Millimoles of \(\mathrm {\mathrm{NaOH}=5 \times 0.4=2}\) Millimoles of \(\mathrm {\mathrm{HCl}=20 \times 0.1=2}\) So, 2 millimoles of \(\mathrm {\mathrm{NaOH}}\) and 2 millimoles of \(\mathrm {\mathrm{HCl}}\) completely neutralise each other, resulting into a neutral solution \(\mathrm {(\mathrm{pH}=7)}\).
CHXI07:EQUILIBRIUM
315000
Determine the \(\mathrm{pH}\) of the solution that results from the addition of \(20.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) to \(30.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{HCl}\).
1 11.30
2 10.55
3 2.70
4 83.35
Explanation:
Millimoles of \(\mathrm{H}^{+}=30 \times 0.01=0.3\) Millimole of \(\mathrm{OH}^{-}=20 \times 0.01 \times 2=0.4\) Remaining millimoles of \(\mathrm{OH}^{-}\) \(\begin{aligned}= & 0.4-0.3=0.1 \\\Rightarrow\left[\mathrm{OH}^{-}\right] & =\dfrac{0.1}{50} \text { or } 2 \times 10^{-3}\end{aligned}\) So, \(\mathrm{pOH}=2.7\) \(\mathrm{pH}=14-2.7=11.30\)
314996
HCOOH and \(\mathrm{CH}_{3} \mathrm{COOH}\) solutions have equal pH . If \(\mathrm{K}_{1} / \mathrm{K}_{2}\) is 4 , their molar concentration ratio will be
1 2
2 0.5
3 4
4 0.25
Explanation:
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{H}^{+}\right]=\mathrm{C} \alpha=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{C}}\) Given pH of two acids as same, so \(\sqrt{\mathrm{K}_{\mathrm{a}_{1}} \mathrm{C}_{1}}=\sqrt{\mathrm{K}_{\mathrm{a}_{2}} \mathrm{C}_{2}}\) \(\dfrac{\mathrm{K}_{\mathrm{a}_{1}}}{\mathrm{~K}_{\mathrm{a}_{2}}}=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) \(4=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) or \(\dfrac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\dfrac{1}{4}=0.25\)
CHXI07:EQUILIBRIUM
314997
When \({\text{50 mL 0}}{\text{.6 M HCl}}\) and \({\text{50 mL 0}}{\text{.3 M NaOH}}\) are mixed, the species with maximum concentration in the resulting solution is
314998
Equal volumes of \(0.1 \mathrm{M}\) potassium hydroxide and \(0.1 \mathrm{M}\) sulphuric acid are mixed. The \(\mathrm{pH}\) of resulting solution is
1 7
2 0
3 Less than 7
4 Greater than 7
Explanation:
m.eq. of acid \(\mathrm {=0.1 \times 2 \times \mathrm{V}}\) m.eq. of base \(\mathrm {=0.1 \times \mathrm{V}}\) Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.
CHXI07:EQUILIBRIUM
314999
\(5 \mathrm{~mL}\) of \({\rm{0}}{\rm{.4\;N}}\,{\rm{NaOH}}\) is mxied with \(20 \mathrm{~mL}\) of \({\rm{0}}{\rm{.1\;N}}\,\,{\rm{HCl}}\) The \(\mathrm{pH}\) of the resulting soluiton will be
1 6
2 7
3 8
4 5
Explanation:
Millimoles of \(\mathrm {\mathrm{NaOH}=5 \times 0.4=2}\) Millimoles of \(\mathrm {\mathrm{HCl}=20 \times 0.1=2}\) So, 2 millimoles of \(\mathrm {\mathrm{NaOH}}\) and 2 millimoles of \(\mathrm {\mathrm{HCl}}\) completely neutralise each other, resulting into a neutral solution \(\mathrm {(\mathrm{pH}=7)}\).
CHXI07:EQUILIBRIUM
315000
Determine the \(\mathrm{pH}\) of the solution that results from the addition of \(20.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) to \(30.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{HCl}\).
1 11.30
2 10.55
3 2.70
4 83.35
Explanation:
Millimoles of \(\mathrm{H}^{+}=30 \times 0.01=0.3\) Millimole of \(\mathrm{OH}^{-}=20 \times 0.01 \times 2=0.4\) Remaining millimoles of \(\mathrm{OH}^{-}\) \(\begin{aligned}= & 0.4-0.3=0.1 \\\Rightarrow\left[\mathrm{OH}^{-}\right] & =\dfrac{0.1}{50} \text { or } 2 \times 10^{-3}\end{aligned}\) So, \(\mathrm{pOH}=2.7\) \(\mathrm{pH}=14-2.7=11.30\)
314996
HCOOH and \(\mathrm{CH}_{3} \mathrm{COOH}\) solutions have equal pH . If \(\mathrm{K}_{1} / \mathrm{K}_{2}\) is 4 , their molar concentration ratio will be
1 2
2 0.5
3 4
4 0.25
Explanation:
\(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{H}^{+}\right]=\mathrm{C} \alpha=\sqrt{\mathrm{K}_{\mathrm{a}} \mathrm{C}}\) Given pH of two acids as same, so \(\sqrt{\mathrm{K}_{\mathrm{a}_{1}} \mathrm{C}_{1}}=\sqrt{\mathrm{K}_{\mathrm{a}_{2}} \mathrm{C}_{2}}\) \(\dfrac{\mathrm{K}_{\mathrm{a}_{1}}}{\mathrm{~K}_{\mathrm{a}_{2}}}=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) \(4=\dfrac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) or \(\dfrac{\mathrm{C}_{1}}{\mathrm{C}_{2}}=\dfrac{1}{4}=0.25\)
CHXI07:EQUILIBRIUM
314997
When \({\text{50 mL 0}}{\text{.6 M HCl}}\) and \({\text{50 mL 0}}{\text{.3 M NaOH}}\) are mixed, the species with maximum concentration in the resulting solution is
314998
Equal volumes of \(0.1 \mathrm{M}\) potassium hydroxide and \(0.1 \mathrm{M}\) sulphuric acid are mixed. The \(\mathrm{pH}\) of resulting solution is
1 7
2 0
3 Less than 7
4 Greater than 7
Explanation:
m.eq. of acid \(\mathrm {=0.1 \times 2 \times \mathrm{V}}\) m.eq. of base \(\mathrm {=0.1 \times \mathrm{V}}\) Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.
CHXI07:EQUILIBRIUM
314999
\(5 \mathrm{~mL}\) of \({\rm{0}}{\rm{.4\;N}}\,{\rm{NaOH}}\) is mxied with \(20 \mathrm{~mL}\) of \({\rm{0}}{\rm{.1\;N}}\,\,{\rm{HCl}}\) The \(\mathrm{pH}\) of the resulting soluiton will be
1 6
2 7
3 8
4 5
Explanation:
Millimoles of \(\mathrm {\mathrm{NaOH}=5 \times 0.4=2}\) Millimoles of \(\mathrm {\mathrm{HCl}=20 \times 0.1=2}\) So, 2 millimoles of \(\mathrm {\mathrm{NaOH}}\) and 2 millimoles of \(\mathrm {\mathrm{HCl}}\) completely neutralise each other, resulting into a neutral solution \(\mathrm {(\mathrm{pH}=7)}\).
CHXI07:EQUILIBRIUM
315000
Determine the \(\mathrm{pH}\) of the solution that results from the addition of \(20.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\) to \(30.00 \mathrm{~mL}\) of \(0.01 \mathrm{M} \mathrm{HCl}\).
1 11.30
2 10.55
3 2.70
4 83.35
Explanation:
Millimoles of \(\mathrm{H}^{+}=30 \times 0.01=0.3\) Millimole of \(\mathrm{OH}^{-}=20 \times 0.01 \times 2=0.4\) Remaining millimoles of \(\mathrm{OH}^{-}\) \(\begin{aligned}= & 0.4-0.3=0.1 \\\Rightarrow\left[\mathrm{OH}^{-}\right] & =\dfrac{0.1}{50} \text { or } 2 \times 10^{-3}\end{aligned}\) So, \(\mathrm{pOH}=2.7\) \(\mathrm{pH}=14-2.7=11.30\)
