CHEMISTRY PH OF BA(OH)2

Question

Assuming complete dissociation, what is the pH of a 3.58mg/L Ba(OH)₂ solution?

Answer 1

The molecular weight of Ba(OH)₂ is 171.34 g/mol. We will determine the concentration of hydroxide anion [OH^-] using dimensional analysis by first converting [Ba(OH)₂] from mg/L to mol/L and then multiplying by 2, because there are 2 moles of OH^- for every mole of Ba(OH)₂ and we are told that it is complete dissociation.

(3.58 mg Ba(OH)₂/L) × (1 mmol Ba(OH)₂ ÷ 171.34 mg Ba(OH)₂ ) × (1 mol Ba(OH)₂ ÷ 1000 mmol Ba(OH)₂ ) × (2 mol OH^- ÷ 1 mol Ba(OH)₂ ) = 4.17 × 10^-5 mol OH^-/L.

Just like pH is defined as the negative logarithm of the hydronium ion concentration (in moles per liter), pOH is the negative logarithm of the hydroxide anion concentration. Therefore:

pOH = -log(4.17 × 10^-5)

pOH = 4.38

The relationship between pOH and pH is

pH + pOH = 14

pH = 14 - 4.38 = 9.62