Question

Calculate the pH of each aqueous solution:

(a) 0.041 M HNO₃;

(b) 0.0041 M NaOH

Answer 1

(a) HNO₃ ---> H⁺ + NO₃^-

1 mole of HNO₃ contains 1 mole of H⁺

So [H⁺] = [HNO₃] = 0.041 M

pH = - log[H⁺]

    = - log 0.041

    = 1.39

(b) 1 mole of NaOH contains 1 mole of OH^-

So [OH^-] = [NaOH] = 0.0041 M

pOH = - log[OH^-]

       = - log 0.0041

      = 2.39

We know that pH + pOH = 14

So pH = 14 - pOH

          = 14 - 2.39

          = 11.61