(a) For 2g of TlOH dissolved in water to give 2 L of solution:

(b) For 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution:

(c) For 0.3 g of NaOH dissolved in water to give 200 mL of solution:

(d) For 1mL of 13.6 M HCl diluted with water to give 1 L of solution:
13.6 × 1 mL = M2 × 1000 mL
(Before dilution) (After dilution)
13.6 × 10–3 = M2 × 1L M2
= 1.36 × 10–2 [H+] = 1.36 × 10–2 pH = – log (1.36 × 10–2 )
= (– 0.1335 + 2) = 1.866 =1.87