Answer:
14.1391 g
Explanation:
The pH of a solution is defined as the negative of the hydrogen ion's concentration.
Thus,
pH = - log [H⁺]
Given that:
pH = 2.85
2.85 = - log [H⁺]
So,
[H⁺] = 0.0014125 M
Also, benzoic acid exists in equilibrium in water as:
C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺ Ka = 6.3×10⁻⁵
The ICE table is:
C₆H₅COOH ⇄ C₆H₅COO⁻ + H⁺
At t = 0 a 0 0
At t = teq a-x x x
The expression for Ka is:
Ka = 6.3×10⁻⁵ = [C₆H₅COO⁻][H⁺] / [C₆H₅COOH]
Also,
6.3×10⁻⁵ = x.x / (a-x)
x = [H⁺] = 0.0014125 M
6.3×10⁻⁵ = (0.0014125)² / (a-0.0014125)
Solving, a = 0.3308 M
Given volume = 350.0 mL = 350.0×10⁻³ L
So, Moles of benzoic acid = Concentration×Volume
Moles = 0.3308×350.0×10⁻³ = 0.11578 moles
Mass = Molar mass × moles
Molar mass of benzoic acid = 122.12 g/mol
So,
Mass = 122.12 g/mol × 0.11578 moles = 14.1391 g
The mass of the benzoic acid involved is 1.3 g.
The pH shows the degree of acidity or alkalinity of the solution. We have been told that the pH of this solution is 2.85. We have the reaction;
C6H5COOH ⇄C6H5COO- + H^+
Ka = [C6H5COO-] [H^+]/[C6H5COOH]
[H^+] = Antilog (-2.85) = 1.4 * 10^-3 M
6.3 x 10-5 = [1.4 * 10^-3]^2/ [C6H5COOH]
[C6H5COOH] = [1.4 * 10^-3]^2/ 6.3 x 10-5
[C6H5COOH] = 1.96 * 10^-6/6.3 x 10-5
= 3.1 * 10^-2 M
number of moles = concentration * volume
number of moles = mass/molar mass
mass/molar mass = concentration * volume
mass = concentration * volume * molar mass
mass = 3.1 * 10^-2 M * 0.35 L * 122 g/mol
mass = 1.3 g
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