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

Question

The pH of a solution obtained by mixing 50 &#160; mL of 1 &#160; M &#160; HCl and 30 &#160; mL of 1 &#160; M &#160; NaOH is x &#215; 10 - 4 . The value of x is (Nearest integer) log &#160; 2 . 5 = 0 . 3979

Aicharya · Accepted Answer

Milli equivalents of HCl N a V a = 50 &#215; 1 = 50 Milli equivalents of NaOH N b V b = 30 &#215; 1 = 30 Since N a V a &#62; 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 &#215; 10 - 1 pH = - log H + = - log 2 . 5 &#215; 10 - 1 = 1 - 0 . 3979 = 0 . 6021 pH &#215; 10 4 = 0 . 6021 &#215; 10 4 = 6021

The pH of a solution obtained by mixing 50   mL of 1   M   HCl and 30   mL of 1   M…

Solution

The pH of a solution obtained by mixing 50 &#160; mL of 1 &#160; M &#160; HCl and 30 &#160; mL of 1 &#160; M…

Solution

The pH of a solution obtained by mixing 50 mL of 1 M HCl and 30 mL of 1 M…