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olchik [2.2K]
3 years ago
11

What is the ph of a solution that contains 1.0 l of 0.10 m ch3cooh and 0.080 m nach3coo after 0.03 moles of naoh added?

Chemistry
1 answer:
zimovet [89]3 years ago
4 0

Answer: pH = 4.996

Explanation:

No of moles = molarity x volume

:• no of moles of CH3COOH = 0.1M x 0.1L

n(CH3COOH) = 0.1mol

Since 0.03mole of NaOH is added, then 0.03 mole of CH3COOH will be converted to the conjugate.

Therefore, Moles of CH3COOH becomes,

0.1 - 0.03 = 0.07 mol

Subsequently, the moles of CH3COONa increases and becomes,

0.08 + 0.03 = 0.11 mol

Using the Hendersom-Hasselbach equation,

pH = pKa + log [Moles of conjugate÷ moles of Ch3COOH]

From literature, pKa of Ch3COOH is 4.8

Thus,

pH = 4.8 + log [0.11/0.07]

pH = 4.8 + 0.1963

pH = 4.996

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Answer:

The concentration of the solution will be much lower than 6M

Explanation:

To prepare a solution of a solid, the appropriate mass is taken and accurately weighed in a weighing balance and then made up to mark with distilled water.

From

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C= 120/40× 1000/500

C=6M

This solution will not be exactly 6M if the student follows the procedure outlined in the question. The actual concentration will be much less than 6M.

This is because, solutions are prepared in a standard volumetric flask. Using a 1000ml beaker, the student must have added more water than the required 500ml thereby making the actual concentration of the solution less than the expected 6M.

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