Question

Calculate analytical concentration for each of the following solutions:

a) HCl solution, pH = 1.34

b) Acetic acid solution, pH = 4.31

c) Sulfuric acid solution, pH = 2.21

d) Potassium hydroxide solution, pH = 12.21

Acetic acid: pKa = 4.76

Sulfuric acid: pKa,1 = strong, pKa,2 = 1.99

In water, [H3O+][OH–] = 1.00 ! 10–14

Ignore the effect of ionic strength.

Answer 1

The pH is then calculated from the [H+ ] via the pH equation. pH = log 1/[H+ ] = - log [H+ ]

From the pH equation, the following relationship can be used to determine the [H+ ]:

[H+ ] = 10^-pH

Again, since the hydrogen ion concentration [H+ ] is equal to the molar concentration of HCl ([H+ ] = mol/L HCl), the concentration of HCl in Wt% can be determined as follows

[H+] = 10^-1.34 = 0.0457 mol/L

HCl in Wt% can be determined as follows = 0.0457 mol/L x 36.46 g/mol = 1.66 g/L

2. HC2H3O2 <-----> H⁺+ C2H3O²⁻

and

Ka = [H+][C2H3O2−] / [HC2H3O2]

So let's do some simplifying. The amount of acetic acid left is the orignial molarity minus the amount that's dissociated. The amount of Hydrogen ion and conjugate base left is equal to the amount of that's been dissociated from the original amount of acid. So let's let HA equal the molarity of the acid and x equal to the amount dissociated. So

Ka = [x] [x] / [HA−x] = x² / HA−x

Now let's solve

Ka (HA−x) = x²

HA−x = x² / Ka

switching the x interms of ph

HA−10^−ph = (10^−ph)² / Ka

plugging in

HA−10^−4.31 = (10^-4.31)² / 1.738 x 10^-5

HA = 1.87 x 10^-4 M

This is the molarity of the acid at equilibrium. Add 10^-pH for the concentration of the acid before dissociation.

10^-4.31 = 4.89 x 10^-5 M = 0.00293 g/L

3. pH = -log [H+]

[H+] = 10^-2.21 = 0.006166 M = 0.6 g/L

4. Since KOH is a strong base, we cannot use pH directly, because it gives the H+ ion concentration. We need OH- concentration.

pH + pOH = 14
pOH = 14 - 12.21 = 1.79

pOH = - log [OH-]

1.79 = - log [OH-]

[OH-] = 0.0162 M

KOH is a strong base and dissociates completely:

KOH(aq) ---------> K+ +(aq) + OH-(aq)

The concentration of KOH is 0.0162 M = 0.9 g/L