Question

Additional Practice Questions

Answer 1

pH of solution = 13.033

Further explanation

Given

2.31 g Ba(OH)₂

250 ml water

Required

pH of solution

Solution

Barium hydroxide is fully ionized, means that Ba(OH)₂ is a strong base

So we use a strong base formula to find the pH

[OH ⁻] = b. Mb where

b = number of OH⁻ /base valence

Mb = strong base concentration

Molarity of Ba(OH)₂(MW=171.34 g/mol) :

$\tt M=\dfrac{mol}{L}=\dfrac{2.31~g\div 171.34~g/mol}{0.25~L}\\\\M=0.054$

Ba(OH)₂ ⇒ Ba²⁺ + 2OH⁻(b=valence=2)

[OH⁻]= 2 . 0.054

[OH⁻] = 0.108

pOH= - log 0.108

pOH=0.967

pOH+pH=14

pH=14-0.967

pH=13.033