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


Answer 1

The mass of the benzoic acid involved is 1.3 g.

What is pH?

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

Learn more about pH:

Answer 2


14.1391 g


The pH of a solution is defined as the negative of the hydrogen ion's concentration.


pH = - log [H⁺]

Given that:

pH = 2.85

2.85 = - log [H⁺]


[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]


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


Mass = 122.12 g/mol × 0.11578 moles = 14.1391 g

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