(a) for 10 mL of 0.2M Ca(OH)₂ and 25 mL of 0.1M HCl

10 ml of 0.2 M Ca(OH)₂= 10× 0.2 milimole = 2 milimoles

25 ml of 0.1 M HCl = 25× 0.1 milimole = 2.5 milimoles

Ca(OH)₂ + 2HCl ↔ CaCl₂ + 2H₂O

0.1 a mole of Ca(OH)₂ reacts with 2 millimole of HCl

∴ 2.5 mole of HCl reacts with 1.25 millimole of Ca(OH)₂

_∴ Ca(OH)₂ left = 1=1.25 = 0.75 milimole

Volume of reaction mixture = 10mL+ 25mL = 35mL

∴ Molarity of Ca(OH)₂in the mixture solution is:

Molarity =

Hence,

Molarity = = 0.0214 M

⇒[OH^-] = 2 × 0.0124 M

⇒[OH^-] = 0.0428 M

To caculate pOH, we use the formula:

pH = - log[OH^-]

⇒pH= - log (4.28× 10^-2)

⇒pH = -log 4.28 + 2log 10

⇒pH = 2 – 0.6314

⇒pH = 1.37

As we know that, pH + pOH =14

∴ pH = 14 – 1.37

⇒pH = 12.63

Hence, the pH is 12.62

(b) for 10 mL of 0.01M H₂SO₄ and 10 mL of 0.01M Ca(OH)₂

10 ml of 0.01 M H₂sO₄ = 10× 0.01 milimole = 0.1 milimole

10 ml of 0.01 M Ca(OH)₂ = 10× 0.01 milimole = 0.1 milimole

Ca(OH)₂ + H₂sO₄ = CaSO₄ + 2H₂O

1 mole of Ca(OH)₂ reacts with 1 mole of H₂sO₄

∴ 0.1 millimole of Ca(OH)₂ will react completely with 0.1 minimal of H₂SO₄. Hence solution will be neutral with pH= 7

(C) 10 mL of 0.1M H₂SO₄ and 10 mL of 0.1M KOH

10 ml of 0.01 M H₂sO₄ = 10× 0.01 milimole = 0.1 milimole

10 ml of 0.1 M KOH = 10× 0.1 milimole = 1 milimole

2KOH+ H₂sO₄ = K₂SO₄ + 2H₂O

I mole of KOHreacts with 0.5 milimole of H₂sO₄

_∴ H₂SO₄ left = 1=0.5 = 0.5 milimole

Volume of reaction mixture = 10+ 10 = 20 mL

∴ Molarity of H₂SO₄ in the mixture solution is:

Molarity =

Hence,

Molarity = = 2.5 × 10^-12 M

⇒[H⁺] = 2 × 2.5 × 10^-12

⇒[H⁺] = 5× 10^-2

To caculate pH, we use the formula:

pH = - log[H⁺]

⇒pH= - log (5× 10^-2)

⇒pH = -log 5 + 2log 10

⇒pH = 2 – 0.699

⇒pH = 1.3

Hence, the pH is 1.3