NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI07:EQUILIBRIUM
315045
Assertion : \(\mathrm{pH}\) of \(10\,{\text{M HCl}}\) aqueous solution is less than 1 . Reason : \(\mathrm{pH}\) is equal to negative logarithm of \(\mathrm{H}^{+}\)concentration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Theoretically, the \(\mathrm{pH}\) should can be negative. \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{1}=-1\) So option (1) is correct.
CHXI07:EQUILIBRIUM
315035
Calculate the \(\mathrm{pH}\) of a \(0.10\,\,{\text{M}}\) ammonia solution. The dissociation constant of ammonia, \({{\rm{K}}_{\rm{b}}}{\rm{ = 1}}{\rm{.77 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)
315037
If the dissociation constant of \(5 \times 10^{-4} \mathrm{M}\) aqueous solution of diethylamine is \(2.5 \times 10^{-5}\), its \(\mathrm{pH}\) value is
1 8
2 3.95
3 10.04
4 2
Explanation:
Diethylamine is a weak base and from Ostwald's dilution law, for weak base, \(\mathrm{K}_{\mathrm{b}}=\mathrm{C} \alpha^{2}\) \(\therefore \alpha=\sqrt{\dfrac{2.5 \times 10^{-5}}{5 \times 10^{-4}}}=0.22\)\(\left[\mathrm{OH}^{-}\right]=\mathrm{C} \alpha=5 \times 10^{-4} \times 0.22\) \(=1.1 \times 10^{-4}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]\) \(=-\log \left(1.1 \times 10^{-4}\right)=3.96\) \(\therefore \mathrm{pH}=14-3.96=10.04\)
315045
Assertion : \(\mathrm{pH}\) of \(10\,{\text{M HCl}}\) aqueous solution is less than 1 . Reason : \(\mathrm{pH}\) is equal to negative logarithm of \(\mathrm{H}^{+}\)concentration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Theoretically, the \(\mathrm{pH}\) should can be negative. \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{1}=-1\) So option (1) is correct.
CHXI07:EQUILIBRIUM
315035
Calculate the \(\mathrm{pH}\) of a \(0.10\,\,{\text{M}}\) ammonia solution. The dissociation constant of ammonia, \({{\rm{K}}_{\rm{b}}}{\rm{ = 1}}{\rm{.77 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)
315037
If the dissociation constant of \(5 \times 10^{-4} \mathrm{M}\) aqueous solution of diethylamine is \(2.5 \times 10^{-5}\), its \(\mathrm{pH}\) value is
1 8
2 3.95
3 10.04
4 2
Explanation:
Diethylamine is a weak base and from Ostwald's dilution law, for weak base, \(\mathrm{K}_{\mathrm{b}}=\mathrm{C} \alpha^{2}\) \(\therefore \alpha=\sqrt{\dfrac{2.5 \times 10^{-5}}{5 \times 10^{-4}}}=0.22\)\(\left[\mathrm{OH}^{-}\right]=\mathrm{C} \alpha=5 \times 10^{-4} \times 0.22\) \(=1.1 \times 10^{-4}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]\) \(=-\log \left(1.1 \times 10^{-4}\right)=3.96\) \(\therefore \mathrm{pH}=14-3.96=10.04\)
315045
Assertion : \(\mathrm{pH}\) of \(10\,{\text{M HCl}}\) aqueous solution is less than 1 . Reason : \(\mathrm{pH}\) is equal to negative logarithm of \(\mathrm{H}^{+}\)concentration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Theoretically, the \(\mathrm{pH}\) should can be negative. \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{1}=-1\) So option (1) is correct.
CHXI07:EQUILIBRIUM
315035
Calculate the \(\mathrm{pH}\) of a \(0.10\,\,{\text{M}}\) ammonia solution. The dissociation constant of ammonia, \({{\rm{K}}_{\rm{b}}}{\rm{ = 1}}{\rm{.77 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)
315037
If the dissociation constant of \(5 \times 10^{-4} \mathrm{M}\) aqueous solution of diethylamine is \(2.5 \times 10^{-5}\), its \(\mathrm{pH}\) value is
1 8
2 3.95
3 10.04
4 2
Explanation:
Diethylamine is a weak base and from Ostwald's dilution law, for weak base, \(\mathrm{K}_{\mathrm{b}}=\mathrm{C} \alpha^{2}\) \(\therefore \alpha=\sqrt{\dfrac{2.5 \times 10^{-5}}{5 \times 10^{-4}}}=0.22\)\(\left[\mathrm{OH}^{-}\right]=\mathrm{C} \alpha=5 \times 10^{-4} \times 0.22\) \(=1.1 \times 10^{-4}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]\) \(=-\log \left(1.1 \times 10^{-4}\right)=3.96\) \(\therefore \mathrm{pH}=14-3.96=10.04\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
CHXI07:EQUILIBRIUM
315045
Assertion : \(\mathrm{pH}\) of \(10\,{\text{M HCl}}\) aqueous solution is less than 1 . Reason : \(\mathrm{pH}\) is equal to negative logarithm of \(\mathrm{H}^{+}\)concentration.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Theoretically, the \(\mathrm{pH}\) should can be negative. \(\mathrm{pH}=-\log \left[\mathrm{H}^{+}\right]=-\log 10^{1}=-1\) So option (1) is correct.
CHXI07:EQUILIBRIUM
315035
Calculate the \(\mathrm{pH}\) of a \(0.10\,\,{\text{M}}\) ammonia solution. The dissociation constant of ammonia, \({{\rm{K}}_{\rm{b}}}{\rm{ = 1}}{\rm{.77 \times 1}}{{\rm{0}}^{{\rm{ - 5}}}}\)
315037
If the dissociation constant of \(5 \times 10^{-4} \mathrm{M}\) aqueous solution of diethylamine is \(2.5 \times 10^{-5}\), its \(\mathrm{pH}\) value is
1 8
2 3.95
3 10.04
4 2
Explanation:
Diethylamine is a weak base and from Ostwald's dilution law, for weak base, \(\mathrm{K}_{\mathrm{b}}=\mathrm{C} \alpha^{2}\) \(\therefore \alpha=\sqrt{\dfrac{2.5 \times 10^{-5}}{5 \times 10^{-4}}}=0.22\)\(\left[\mathrm{OH}^{-}\right]=\mathrm{C} \alpha=5 \times 10^{-4} \times 0.22\) \(=1.1 \times 10^{-4}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]\) \(=-\log \left(1.1 \times 10^{-4}\right)=3.96\) \(\therefore \mathrm{pH}=14-3.96=10.04\)