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natta225 [31]
3 years ago
14

The reactant which is consumed completely in a reaction is called the excess reactant. TRUE or FALSE​

Chemistry
1 answer:
lina2011 [118]3 years ago
5 0

Answer:

False

Please rate brainliest

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Answer:

Energy is transferred from one object to another when a reaction takes place.

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3 years ago
Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1
givi [52]

According to Hasselbach-Henderson equation:

pH=pK_{a}+log\frac{[A^{-}]}{[HA]}

Here, [A^{-}] is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is CH_{3}COOK and acid is CH_{3}COOH thus, Hasselbach-Henderson equation will be as follows:

pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{2}{1}=5.06

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{3}=4.28

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{5}{1}=5.45

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{1}=4.76

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}

Also, pK_{a}=4.76

Putting the values in equation (1),

pH=4.76+log\frac{1}{10}=3.76

Therefore, pH of solution is 3.76.

4 0
3 years ago
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