acid concentration

Question

An unknown amount of acid, H2A, was dissolved in pure water, so the volume of the solution became 100 ml, and the pH was 3.7, If the Ka of the acid is 1×10 ^-7, and the molar mass of the acid is 34 g/mol, find the initial concentration of the acid, and the mass of the dissolved acid

Answer 1

H₂A <==> H+ + HA- Ka1 = 1x10^-7

HA- <==> H+ + A^2- Ka2 = not given (assume it is quite small and insignificant)

Ka = 1x10^-7 = [H⁺][HA^-] / [H₂A]

From the pH = 3.7, we can find the [H⁺] and [HA^-]

[H⁺] = [HA^-] = 1x10^-3.7

[H⁺] = [ HA^-] = 1.995x10^-4 M

1x10^-7 = (1.995x10^-4 M)(1.995x10^-4 M) / [H₂A] - 1.995x10^-4 M (assume 1.995x10^-4 M is small)

1x10^-7 = (1.995x10^-4 M)² / [H₂A]

[H₂A] = 0.398 M = 0.040 M (2 significant figures) (assumption was valid as 1.995x10^-4 is small relative to 0.398)

Mass of dissolved acid: 0.04 M = 0.04 moles/L x 0.100 L = 0.040 moles of dissolved acid

0.040 moles acid x 34 g / mol = 1.4 g of acid dissolved (2 significant figures)

Answer 2

pH = -log[H+]

3.7 = -log[H+]

[H+] = 10^-3.7

Therefore…Hydrogen ion concentration = 2x10^-4 M when the unknown quantity of our acid H2A is diluted to a 100ml volume.

H2A -> H+ + HA-

I. ?

C. -x +x +x

Eq. ?-x x x

the hydrogen ion concentration [H+] from the equilibrium expression for the Ka equation is that found from the pH equation.

Ka = productcs^coefficients / reactants^coefficients

using,

Ka = [H+][HA-] / [H2A]

at equillibrium,

1x10^-7 = [2x10^-4][2x10-4]/[?-2x10^-4]

Let us solve for the question mark, which represents the initial morality of the H2A acid diluted.

? = [2x10^-4]^2/(1x10^-7) + 2x10^-4

? = 0.4002M

H2A = 0.4002M at 100ml dilution

0.4002 mole/L x 0.1L = 0.04002 mole H2A is the initial concentration of the acid

0.04002 mole x 34 g/mole = 1.36 grams is the mass of the H2A acid dissolved.