What mass of benzoic acid would you dissolve in 350.0 ml of water to produce a solution with a ph = 2.85?

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

14 days is the half-life of phosphorus-32

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