JEE Main202131 Aug 2021Evening ShiftChemistryIonic EquilibriumActual
The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M NaOH is x × 10 - 4 . The value of x is (Nearest integer) log 2 . 5 = 0 . 3979
Correct answer
0
Step-by-step solution
Milli equivalents of HCl N a V a = 50 × 1 = 50 Milli equivalents of NaOH N b V b = 30 × 1 = 30 Since N a V a > N b V b H + = N a V a - N b V b V a + V b = 50 - 30 80 = 20 80 = 0 . 25 = 2 . 5 × 10 - 1 pH = - log H + = - log 2 . 5 × 10 - 1 = 1 - 0 . 3979 = 0 . 6021 pH × 10 4 = 0 . 6021 × 10 4 = 6021