SATHEE: Equilibrium Question 886

Equilibrium Question 886

Question: The ionization constant of $H C O O H$ is $1.8 \times 10^{- 4}$ . What is the percent ionization of a 0.001 M solution'

Options:

A) 66%

B) 42%

C) 34%

D) 58%

Show Answer

Answer:

Correct Answer: C

Solution:

$\underset{0.001 - x}{H C O O H} {\underset{x}{H C O O},}^{-} + {\underset{x}{H},}^{+}$

As $α = \sqrt{\frac{K_{a}}{C}} = \sqrt{\frac{1.8 \times 10^{- 4}}{0.001}} = 0.42$

As $α > 0.1$ so can’t be neglected, $1.8 \times 10^{- 4} = \frac{x^{2}}{0.001 - x}$ and $x = 3.4 \times 10^{- 4}$

Now, % ionization $= \frac{I o n i s e d, H C O O H}{T o t a l, H C O O H} = \frac{3.4 \times 10^{- 4}}{0.001} \times 100 = 34$ %

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Equilibrium Question 886

Question: The ionization constant of HCOOH is 1.8×10−4 . What is the percent ionization of a 0.001 M solution'

Options:

Answer:

Solution:

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Question: The ionization constant of $H C O O H$ is $1.8 \times 10^{- 4}$ . What is the percent ionization of a 0.001 M solution'