04. Hydrogen Ion Concentration - pH Scale
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Ionic Equilibrium

229814 At $100^{\circ} \mathrm{C}$ the $\mathrm{K}_w$ of water is 55 times its value at $25^{\circ} \mathrm{C}$, What will be the $\mathrm{pH}$ of neutral solution? (log $55=1.74)$

1 7.00
2 7.87
3 5.13
4 6.13
Karnataka NEET-2013
Ionic Equilibrium

229816 $\mathrm{K}_{\mathrm{p}}$ for the following reaction will be equal to $\mathrm{Fe}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \square \quad \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{H}_2(\mathrm{f})$

1 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4\left(\mathrm{p}_{\mathrm{F}_2, \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{Fo}}\right)^3\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
2 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)}$
3 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
4 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)\left(\mathrm{p}_{\mathrm{Fe}_4 \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{F}}\right)}$
COMEDK-2016
Ionic Equilibrium

229817 The highest $\mathrm{pH}$ is exhibited by

1 $0.001 \mathrm{M} \mathrm{KOH}$
2 $0.01 \mathrm{M} \mathrm{KOH}$
3 $0.1 \mathrm{M} \mathrm{HCl}$
4 $0.01 \mathrm{M} \mathrm{HCl}$
COMEDK-2017
Ionic Equilibrium

229818 $5 \mathrm{~L}$ of $\mathrm{NaOH}$ solution of $\mathrm{pH} 12$ contains

1 $200 \mathrm{~g}$
2 $0.2 \mathrm{~g}$
3 $20 \mathrm{~g}$
4 $2 \mathrm{~g}$
COMEDK-2020
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Ionic Equilibrium

229814 At $100^{\circ} \mathrm{C}$ the $\mathrm{K}_w$ of water is 55 times its value at $25^{\circ} \mathrm{C}$, What will be the $\mathrm{pH}$ of neutral solution? (log $55=1.74)$

1 7.00
2 7.87
3 5.13
4 6.13
Karnataka NEET-2013
Ionic Equilibrium

229816 $\mathrm{K}_{\mathrm{p}}$ for the following reaction will be equal to $\mathrm{Fe}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \square \quad \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{H}_2(\mathrm{f})$

1 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4\left(\mathrm{p}_{\mathrm{F}_2, \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{Fo}}\right)^3\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
2 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)}$
3 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
4 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)\left(\mathrm{p}_{\mathrm{Fe}_4 \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{F}}\right)}$
COMEDK-2016
Ionic Equilibrium

229817 The highest $\mathrm{pH}$ is exhibited by

1 $0.001 \mathrm{M} \mathrm{KOH}$
2 $0.01 \mathrm{M} \mathrm{KOH}$
3 $0.1 \mathrm{M} \mathrm{HCl}$
4 $0.01 \mathrm{M} \mathrm{HCl}$
COMEDK-2017
Ionic Equilibrium

229818 $5 \mathrm{~L}$ of $\mathrm{NaOH}$ solution of $\mathrm{pH} 12$ contains

1 $200 \mathrm{~g}$
2 $0.2 \mathrm{~g}$
3 $20 \mathrm{~g}$
4 $2 \mathrm{~g}$
COMEDK-2020
NEET DIGITAL LIBRARY
Ionic Equilibrium

229814 At $100^{\circ} \mathrm{C}$ the $\mathrm{K}_w$ of water is 55 times its value at $25^{\circ} \mathrm{C}$, What will be the $\mathrm{pH}$ of neutral solution? (log $55=1.74)$

1 7.00
2 7.87
3 5.13
4 6.13
Karnataka NEET-2013
Ionic Equilibrium

229816 $\mathrm{K}_{\mathrm{p}}$ for the following reaction will be equal to $\mathrm{Fe}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \square \quad \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{H}_2(\mathrm{f})$

1 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4\left(\mathrm{p}_{\mathrm{F}_2, \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{Fo}}\right)^3\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
2 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)}$
3 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
4 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)\left(\mathrm{p}_{\mathrm{Fe}_4 \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{F}}\right)}$
COMEDK-2016
Ionic Equilibrium

229817 The highest $\mathrm{pH}$ is exhibited by

1 $0.001 \mathrm{M} \mathrm{KOH}$
2 $0.01 \mathrm{M} \mathrm{KOH}$
3 $0.1 \mathrm{M} \mathrm{HCl}$
4 $0.01 \mathrm{M} \mathrm{HCl}$
COMEDK-2017
Ionic Equilibrium

229818 $5 \mathrm{~L}$ of $\mathrm{NaOH}$ solution of $\mathrm{pH} 12$ contains

1 $200 \mathrm{~g}$
2 $0.2 \mathrm{~g}$
3 $20 \mathrm{~g}$
4 $2 \mathrm{~g}$
COMEDK-2020
Join With Us for More Updates
Ionic Equilibrium

229814 At $100^{\circ} \mathrm{C}$ the $\mathrm{K}_w$ of water is 55 times its value at $25^{\circ} \mathrm{C}$, What will be the $\mathrm{pH}$ of neutral solution? (log $55=1.74)$

1 7.00
2 7.87
3 5.13
4 6.13
Karnataka NEET-2013
Ionic Equilibrium

229816 $\mathrm{K}_{\mathrm{p}}$ for the following reaction will be equal to $\mathrm{Fe}(\mathrm{s})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \square \quad \mathrm{Fe}_3 \mathrm{O}_4(\mathrm{~s})+\mathrm{H}_2(\mathrm{f})$

1 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4\left(\mathrm{p}_{\mathrm{F}_2, \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{Fo}}\right)^3\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
2 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)}$
3 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)^4}{\left(\mathrm{p}_{\mathrm{H}_2 \mathrm{O}}\right)^4}$
4 $\frac{\left(\mathrm{p}_{\mathrm{H}_2}\right)\left(\mathrm{p}_{\mathrm{Fe}_4 \mathrm{O}_4}\right)}{\left(\mathrm{p}_{\mathrm{F}}\right)}$
COMEDK-2016
Ionic Equilibrium

229817 The highest $\mathrm{pH}$ is exhibited by

1 $0.001 \mathrm{M} \mathrm{KOH}$
2 $0.01 \mathrm{M} \mathrm{KOH}$
3 $0.1 \mathrm{M} \mathrm{HCl}$
4 $0.01 \mathrm{M} \mathrm{HCl}$
COMEDK-2017
Ionic Equilibrium

229818 $5 \mathrm{~L}$ of $\mathrm{NaOH}$ solution of $\mathrm{pH} 12$ contains

1 $200 \mathrm{~g}$
2 $0.2 \mathrm{~g}$
3 $20 \mathrm{~g}$
4 $2 \mathrm{~g}$
COMEDK-2020
For More Updates join with Us