Solve the f

1. In distilled water: [H₃O ⁺ ] × [OH^-] = 10^-14

For solution A, [H₃O ⁺ ] = 10^-14 ÷ [OH^-]

= 10^-14 ÷ (4.3 × 10^-4)

= 2.32 × 10^-11 M

For solution B, [H₃O ⁺ ] = 7.3 × 10^-10 M

•Since the hydronium ion concentration of solution B is more, so its pH will be less.

•The solution having the more pH will be more basic. As solution A will have more pH so it will be more basic.

2. using the relation, pOH + pH = 14

pOH = 14 – pH

= 14 – 9.3

= 4.7

Using, pOH = -log₁₀[OH^-]

4.7 = -log₁₀[OH^-]

log₁₀[OH^-] = -4.7

log₁₀[OH^-] = 5 -4.7 -5

log₁₀[OH^-] = -5 + 0.3

log₁₀[OH^-] = + 0.3

[OH^-] = antilog

[OH^-] = 1.99 × 10^-5

3. molar mass of NaOH = atomic mass of Na + Atomic mass of O + atomic mass of H = 23 + 16 + 1 = 40g/ mol.

Volume of solution = 5 litres

Given weight = 8g

No. of moles = given weight ÷ molar mass

= 8g ÷ 40 g/mol

= 0.2 mole

Molarity = no. of moles ÷ volume in litres

= 0.2 ÷ 5

= 0.04 M

Therefore, the hydroxide ion concentration [OH^-] = 0.04 M

pOH = -log₁₀[OH^-]

pOH = -log₁₀[ 4 × 10^-2]

pOH = 2 -log₁₀[ 4]

pOH = 2- 0.6020 = 1.39

using, pOH + pH = 14

pH = 14 – pOH

pH = 14 - 1.39 = 12.61