Question

1. Calculate the pHof each of the following solutions of a strong acid in water a. 0.10 M HCI b. 1.0x 10-11 M HCI c. 5.0 M HCI 2. What are the major species present in 0.250 M solutions of cach of the following acids? Calculate the pH of each of these solutions. a. HNO, b. CH COH (HC;H3O;) 3. Calculate the concentration of all species present and the pH of a 0.020 M HF solution.

Answer 1

1.

pH = - log[H₃O⁺]

Strong acid completely ionises into Hydronium ion

Hence,

a.

pH = - log(0.10) = 1.

b.

As concentration of HCl is very low ionisation of water has to be considered.

So, [H₃O⁺] = 1.0×10^-11 + 10^-7 = 10^-7 × ( 1.0+ 0.0001)

= 1.0001×10^-7 M.

Now, pH = - log (1.0001×10^-7) = 6.99.

c.

pH = - log (5.0) = 0.6989.

2.

a.

In weak acid major species is unionized acid molecule

Major species in HNO₂ ; HNO₂  

Minor species : H₃O⁺, NO₂^- .

pH = $\frac{1}{2}$ ( pKa - logC)

= $\frac{1}{2}$ ( 3.15 - log 0.250)

= 1.88

b.

Major species in CH₃COOH ; CH₃COOH

Minor species : H₃O⁺ , CH₃COO^- .

pH = $\frac{1}{2}$ (pKa - logC)

= $\frac{1}{2}$ (4.75 - log 0.250)

= 2.68.

3.

HF is weak acid.

ICE table is

	HF	H3O+	F-
I	C	0	0
C	- Cx	Cx	Cx
E	C - Cx	Cx	Cx

For weak acid , x is very less ( x is degree of dissociation)

So, Ka = Cx²

Ka of HF = 6.8×10^-4

So, x² = (Ka/C) = (6.8×10^-4/0.020) = 0.034

Or, x = 0.184.

So, [H₃O⁺] = Cx = 0.020× 0.184 = 0.0037 M.

[F^-] = Cx = 0.0037 M.

[HF] = C - Cx = 0.020 - 0.0037 = 0.0163 M.

So,

pH = - log [H₃O⁺]

= log(0.0037)

= 2.43