ph of the following chp 10

Question

What is the pHpH of a solution prepared by dissolving 3.40 gg Ca(OH)2Ca(OH)2 in water to make 815 mLmL of solution?

how many moles of HNO3HNO3 are present if 8.60×10⁻² molmol of Ba(OH)2Ba(OH)2 was needed to neutralize the acid solution?

For example, a solution containing 1 molmol of H2SO4H2SO4 contains 2 molmol of ionizable hydrogen atoms, and would therefore require 2 molmol of NaOHNaOH for neutralization.

Answer 1

molar mass Ca(OH)₂ = 74.09 g / mol

mols Ca(OH)₂ present = 3.40 g x 1 mol / 74.09 g = 0.04589 mols

Volume = 815 ml x 1 L / 1000 ml = 0.815 L

[Ca(OH)₂] = 0.04589 mol / 0.815 L = 0.05631 M

pOH = -log 0.05631

pOH = 1.25

pH = 14 - 1.25

pH = 12.75

The reaction taking place in part 2 is...

2HNO₃ + Ba(OH)₂ ==> Ba(NO₃)₂ + 2H₂O (note it takes 2 mol HNO₃ for each mol Ba(OH)₂)

mols HNO3 = 8.60x10^-2 mol Ba(OH)₂ x 2 mol HNO₃ / mol Ba(OH)₂ = 0.172 mol Ba(OH)₂