Question

1.) If 20.0 mL of glacial acetic acid (pure HC2H3O2) is diluted to 1.20 L with water, what is the pH of the resulting solution? The density of glacial acetic acid is 1.05 g/mL.

2.) For each of the following strong base solutions, determine [OH−],[H3O+], pH, and pOH.

1.15×10⁻² M Ba(OH)2 Express your answer using three significant figures. Enter your answers numerically separated by commas.

[OH−],[H3O+] = ??? M

pH,pOH = ??? M

Answer 1

Given

density = 1.05g/mL.

Volume of glacial acetic acid = 20 ml

Mass = volume * density = 1.05 g/ml * 20 ml = 21 g of acetic acid

Molar mass of acetic acid = 60 g/mol

No. of moles = mass /molar mass = 21 grams / 60 g/mol = .35 mol acetic acid

writing disassociation equation for acetic acid

HC2H3O2 <=> H+ + C2H3O2-

writing formula for Ka

Ka = [H+] [C2H3O2-] / [HC2H3O2]

[HC2H3O2] = No. of moles / Volume of solution made = 0.35 moles /1.2 L = 0.292 mol/L

This is the initial concentration
Make an ICE chart With the initial change and equilibrium values.

[HC2H3O2]   [H+] [C2H3O2-]

I 0.292 0 0

C -x x x

E 0.292-x x x

Ka = 1.8 x 10^-5 = (x) (x) / (0.292 - x)

[H+] = x = 0.00228

pH = -log ( [H+]) = -log(0.00228) = 2.64

pH = 2.64
We find pH by plugging it into log.
-Log(1.77x10^-3) = 2.75 pH