4 ml of HCl solution of pH = 2 is mixed with 6 ml of NaOH solution of pH = 12. What would be the final pH of solution? (log 2 = 0.3)

Step-by-step solution

p H = 2 , H C l = 10 - 2 M & p O H = 2 o r N a O H = 10 - 2 M 4 m l o f 10 - 2 M H C l ≡ 4 × 10 - 5 m o l e s H C l . 6 m l o f 10 - 2 M N a O H ≡ 6 × 10 - 5 m o l e s N a O H After mixing total moles of OH − = moles of NaOH − moles of HCl = 6 × 10 − 5 − 4 × 10 − 5 A f t e r m i x i n g e x c e s s m o l e s o f O H - = 2 × 10 - 5 Total volume = 6 + 4 = 10 ml = 10 × 10 − 3 litre O H - = 2 × 10 - 5 10 × 10 3 = 2 × 10 - 3 o r p O H = 3 - log ⁡ 2 = 3 - 0.3 = 2.7 o r p H = 11.3