The acid-dissociation constant, Ka, for benzoic acid is 6.5 × 10-5. Which will you use to calculate the base-dissociation constant, Kb, for the conjugate base of benzoic acid?

Answers

Answer 1
Answer:

Answer:

The base-dissociation constant, Kb,for the conjugate base of benzoic acid is :

K_(b)=1.54* 10^(-10)}

Explanation:

The product of acid dissociation constant and base dissociation constant is equal to the water dissociation constant.The general formula for the reaction is:

K_(w)=K_(a)K_(b)

For the acid dissociation reaction:

HA + H_(2)O\rightleftharpoons H^(+)+A^(-)

The conjugate base for the acid is A-

The acid is HA . and its Ka is given.

The value of Kw is fixed  at a given  temperature , which is equal to:

K_(w)=10^(-14)

K_(a)=6.5* 10^(-5)

K_(b)=(10^(-14))/(6.5* 10^(-5))

K_(b)=1.54* 10^(-10)}

Answer 2
Answer:

Answer:

Ka*kb=kw

Explanation:Got it right


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A saturated solution of NaNO3 is prepared at 60.°C using 100. grams of water. As this solution is cooled to 10.°C, NaNO3 precipitates (settles)out of the solution. The resulting solution is saturated. Approximately how many grams of NaNO3 settled out of the original solution?
(1) 46 g (3) 85 g
(2) 61 g (4) 126 g

Answers

The answer is 46 g


The explanation :


According to the attached curve of the relation between the temperature and the grams in solutions:


-we can see from the curve that the grams of NaNO3 at 60°C = 126 g


-and the grams of NaNO3 at 10°C = 80 g


-so, To get the grams of NaNO3 settled out of the original solution, we will subtract the grams of NaNO3 at 10°C - the grams of NaNO3 at 60°C


126 g - 80 g = 46 g


∴ the correct answer is 46 g

The answer is (1) 46g. This is related to the solubility curve of the NaNO3. The solubility of NaNO3 under 60 ℃ is 126 g. And the solubility under 10 ℃ is 80 g. So the NaNO3 precipitates is 126-80=46 g.

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