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3 years ago

What is the ph of a solution of 0.50 m acetic acid?

1 answer:

6 0

You need to use the Ka for the acetic acid and the equilibrium equation.

Ka = 1.85 * 10^ -5

Equilibrium reaction: CH3COOH (aq) ---> CH3COO(-) + H(+)

Ka = [CH3COO-][H+] / [CH3COOH]

Molar concentrations at equilibrium

CH3COOH CH3COO- H+

0.50 - x x x

Ka = x*x / (0.50 - x) = x^2 / (0.50 - x)

Given that Ka is << 1 => 0.50 >> x and 0.50 - x ≈ 0.50

=> Ka ≈ x^2 / 0.50

=> x^2 ≈ 0.50 * Ka = 0.50 * 1.85 * 10^ -5 = 0.925 * 10^ - 5 = 9.25 * 10 ^ - 6

=> x = √ [9.25 * 10^ -6] = 3.04 * 10^ -3 ≈ 0.0030

pH = - log [H+] = - log (x) = - log (0.0030) = 2.5

Answer: 2.5

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When 0.250 moles of KCl are added to 200.0 g of water in a constant pressure calorimeter a temperature change is observed. Given

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Explanation:

Upon dissolution of KCl heat is generated and temperature of the solution raises.

Therefore, heat generated by dissolving 0.25 moles of KCl will be as follows.

$17.24 kJ/mol \times 0.25 mol$

= 4.31 kJ

or, = 4310 J (as 1 kJ = 1000 J)

Mass of solution will be the sum of mass of water and mass of KCl.

Mass of Solution = mass of water + (no. of moles of KCl × molar mass)

= 200 g + $(0.25 mol \times 54.5 g/mol)$

= 200 g + 13.625 g

= 213.625 g

Relation between heat, mass and change in temperature is as follows.

Q = $mC \Delta T$

where, C = specific heat of water = $4.184 J/g^{o}C$

Therefore, putting the given values into the above formula as follows.

Q = $mC \Delta T$

4310 J = $213.625 g \times 4.184 J/g^{o}C \times \Delta T$

$\Delta T = 4.82^{o}C$

Thus, we can conclude that rise in temperature will be $4.82^{o}C$ .

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