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2 years ago

What is the concentration of a solution made by dissolving 0.99mol of acetic acid in enough water to make 0.58L of solution?

Chemistry

1 answer:

marissa [1.9K]2 years ago

4 0

Answer:

[CH₃COOH] = 1.70 M

Explanation:

When we talk about concentration we can determine Molarity

Molarity determines the moles of solute that are contained in 1L of solution.

In this case our solute is the acetic acid.

M = mol/L

M = 0.99 mol /0.58L → 1.70 M

We can also make a rule of three

In 0.58 L of solution we have 0.99 moles of solute

In 1 L of solution we may have (1 . 0.99) / 0.58 = 1.70 moles

Acetic acid is a weak acid, partially dissociated in water.

CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka

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erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

Solution : Given,

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Volume of solution = 380 ml = $380\times 10^{-3}$ $(1ml=10^{-3}L)$

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Now we have to calculate the moles of $OH^-$ .

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$\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles$

As, $H_2SO_4\rightarrow 2H^++SO^{2-}_4$

From this reaction, we conclude that

2 moles of $H^+$ ion is given by the 1 mole of $H_2SO_4$

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Therefore, the mass of sulfuric acid needed is $16.23\times 10^{-5}g$ .

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