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lara [203]
3 years ago
9

Guys please help me i really need help! pls and thank you!​

Chemistry
1 answer:
LuckyWell [14K]3 years ago
4 0

Answer:

Hydrogen

Explanation:

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Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH3COO, to 23 g of acetic acid, CH3COOH, and dissolving
Elza [17]

This is an incomplete question, here is a complete question.

Calculate the pH of a solution made by adding 59 g of sodium acetate, NaCH₃COO, to 23 g of acetic acid, CH₃COOH, and dissolving in water to make 400. mL of solution. Hint given in feedback. The Ka for CH₃COOH is 1.8 x 10⁻⁵ M. As usual, report pH to 2 decimal places.

Answer : The pH of the solution is, 4.97

Explanation :

First we have to calculate the moles of CH₃COOH and NaCH₃COO

\text{Moles of }CH_3COOH=\frac{\text{Given mass }CH_3COOH}{\text{Molar mass }CH_3COOH}=\frac{23g}{60g/mol}=0.383mol

and,

\text{Moles of }CH_3COONa=\frac{\text{Given mass }CH_3COONa}{\text{Molar mass }CH_3COOH}=\frac{59g}{82g/mol}=0.719mol

Now we have to calculate the value of pK_a.

The expression used for the calculation of pK_a is,

pK_a=-\log (K_a)

Now put the value of K_a in this expression, we get:

pK_a=-\log (1.8\times 10^{-5})

pK_a=5-\log (1.8)

pK_a=4.7

Now we have to calculate the pH of the solution.

Using Henderson Hesselbach equation :

pH=pK_a+\log \frac{[Salt]}{[Acid]}

pH=pK_a+\log \frac{[CH_3COONa]}{[CH_3COOH]}

Now put all the given values in this expression, we get:

pH=4.7+\log (\frac{0.719}{0.383})

(As the volume is same. So, we can write concentration in terms of moles.)

pH=4.97

Therefore, the pH of the solution is, 4.97

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(3) The nucleus is negative and the electron cloud is positive.

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