KCET2012ChemistryIonic Equilibrium
The pH of the solution obtained by mixing 100 ml of a solution of pH =3 with 400 ~mL of a solution of pH =4 is
Options
- A7- 2.8
- B4- 2.8
- C5- 2.8
- D3- 2.8
Correct answer
B. 4- 2.8
Step-by-step solution
For solution I , pH =3 [ H ⁺ ]=10⁻³ Concentration of solution M ₁=10⁻³ M V ₁=100 ~mL For solution II, pH = 4 [ H ⁺ ]=10⁻⁴ Concentration of solution, M ₂=10⁻⁴ M V ₂=400 ~mL Concentration of resulting solution aligned M &= M ₁ ~V ₁+ M ₂ ~V ₂ ~V ₁+ V ₂ &= 10⁻³ 100+10⁻⁴ 400 100+400 &= 0.14 500 M &=0.00028=2.8 10⁻⁴ [ [ H ⁺ ] . &=2.8 10⁻⁴ pH of resulting solution &=- [ H ⁺ ] &=- (2.8 10⁻⁴ ) &=- 2.8- 10⁻⁴ &=4- 2.8 aligned