SATHEE: Equilibrium Question 314

Equilibrium Question 314

Question: Calculate the amount of ${(N H_{4})}_{2} S O_{4}$ in grams which must be added to 500 ml of $0.200, M, N H_{3}$ to yield a solution with $p H = 9.35$

$(K_{b}$ for $N H_{3} = 1.78 \times 10^{- 5})$ [UPSEAT 2001]

Options:

A) 10.56 gm

B) 15 gm

C) 12.74 gm

D) 16.25 gm

Show Answer

Answer:

Correct Answer: A

Solution:

$p O H = p K b + \log \frac{[S a l t]}{[B a s e]}$

$14 - 9.35 = - \log (1.78 \times 10^{- 5}) + \log \frac{[S a l t]}{100}$

$[S a l t] = 79.9$

Þ $\frac{w}{132} \times 1000 = 79.9$

Þ $w = 10.56$

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Equilibrium Question 314

Question: Calculate the amount of ${(N H_{4})}_{2} S O_{4}$ in grams which must be added to 500 ml of $0.200, M, N H_{3}$ to yield a solution with $p H = 9.35$

Options:

Answer:

Solution:

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Equilibrium Question 314

Question: Calculate the amount of (NH4)2SO4 in grams which must be added to 500 ml of 0.200,M,NH3 to yield a solution with pH=9.35

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: Calculate the amount of ${(N H_{4})}_{2} S O_{4}$ in grams which must be added to 500 ml of $0.200, M, N H_{3}$ to yield a solution with $p H = 9.35$