A solution is prepared by mixing 0 . 01 mol each of H 2 CO 3 , NaHCO 3 , Na 2 CO 3 , and NaOH in 100 ~mL of water. pH of the resulting solution is [Given: pK a 1 and pK a 2 of H 2 CO 3 are 6 . 37 and 10 . 32 , respectively. log 2 = 0 . 30 ]
A. A
First acid base reaction between H 2 CO 3 and NaOH takes place. H 2 CO 3 0 . 01 mole + NaOH 0 . 01 mole ⟶ NaHCO 3 - 0 . 01 mole + H 2 O After the acid base reaction, we have 0 . 01 mole   Na 2 CO 3 and 0 . 02 moles of NaHCO 3 . Here, This will form an acidic buffer of NaHCO 3 and Na 2 CO 3 . ∴   pH = pK a 2 + log Salt Acid = 10 . 32 + log 0 . 01 0 . 1 0 . 02 0 . 1 = 10 . 32 + log 1 2 = 10 . 32 - log 2 = 10 . 32 - 0 . 3 = 10 . 02 ∴   pH = 10 . 02
Related: Chemistry — Ionic Equilibrium · All PYQ Banks