Percentage ionization of water at certain temperature is 3 . 6 × 10 - 7 % . Calculate K w and pH of water.

The density of H 2 O = 1 g / mL i.e., 1 mL water has the mass = 1 g Molarity of water = mass of H 2 O Molar mass of H 2 O × 1000 volume of H 2 O in mL Molarity of water = 1 18 × 1000 1 = 1000 18 M = 55 . 5 M .. Given that percentage ionization of water = 3 . 6 × 10 - 7 % Hence, the degree of ionization of water at the given temperature α = 3 . 6 × 10 - 9 H 2 O ⇌ H + + OH - 55 . 5 - 55 . 5 α 55 . 5 α 55 . 5 α Now the ionic product of water = H + OH - at any temperature K W = ( 55 . 5 α ) 2 K W = 1000 18 2 3 . 6 × 10 - 9 2 = 4 × 10 - 14 M 2 H + = 55 . 5 α H + = 1000 18 × 3 . 6 × 10 - 9 H + = 2 × 10 - 7 M pH = 7 - log 2 = 6 . 7

Percentage ionization of water at certain temperature is 3 . 6 × 10 - 7 % . Calculate K w and pH of water.

The density of H 2 O = 1 g / mL i.e., 1 mL water has the mass = 1 g Molarity of water = mass of H 2 O Molar mass of H 2 O × 1000 volume of H 2 O in mL Molarity of water = 1 18 × 1000 1 = 1000 18 M = 55 . 5 M .. Given that percentage ionization of water = 3 . 6 × 10 - 7 % Hence, the degree of ionization of water at the given temperature α = 3 . 6 × 10 - 9 H 2 O ⇌ H + + OH - 55 . 5 - 55 . 5 α 55 . 5 α 55 . 5 α Now the ionic product of water = H + OH - at any temperature K W = ( 55 . 5 α ) 2 K W = 1000 18 2 3 . 6 × 10 - 9 2 = 4 × 10 - 14 M 2 H + = 55 . 5 α H + = 1000 18 × 3 . 6 × 10 - 9 H + = 2 × 10 - 7 M pH = 7 - log 2 = 6 . 7

The hydroxide ion concentration of a solution is 1 × 10 - 8 M . What is the pH of the solution?

The concentration of one mole of hydroxide ions is 1 × 10 - 11 M The hydroxide ion concentration [ OH - ] = 1 . 0 × 10 - 8 M pOH = - log [ OH - ] = log 1 . 0 × 10 - 8 pH + pOH = 14 pH = 14 - p OH = 14 - 8 = 6 Hence, the pH of the solution is 6 .

The hydroxide ion concentration of a solution is 1 × 10 - 8 M . What is the pH of the solution?

The concentration of one mole of hydroxide ions is 1 × 10 - 11 M The hydroxide ion concentration [ OH - ] = 1 . 0 × 10 - 8 M pOH = - log [ OH - ] = log 1 . 0 × 10 - 8 pH + pOH = 14 pH = 14 - p OH = 14 - 8 = 6 Hence, the pH of the solution is 6 .