Question

If 20.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.70 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

Answer 1

If 20.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.70 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

a)

mass of acid = D*V = 20*1.05 = 21 g

mol of acid = mass/MW = 21/60 = 0.35 mo of acid

[Acid] = mol/V= 0.35 / 1.7 = 0.20588

First, assume the acid:

CH3COOH

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = M;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = M - x

substitute in Ka

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

Ka = x*x/(M-x)

x^2 + Kax - M*Ka = 0

if M = vM; then

x^2 + (1.8*10^-5)x - 0.20588*(1.8*10^-5) = 0

solve for x

x =0.01434

substitute

[H+] = 0 + 0.001916= 0.001916M

pH = -log(H+) = -log(0.001916) = 2.72