Answer:
e = 0.250 M calcium hydroxide
it will produce 0.5 M [OH⁻].
Explanation:
A = 0.100 M HCl
HCl is an acid it would produce negligible amount of OH⁻.
B = 0.100 M magnesium hydroxide
Mg(OH)₂ + H₂O → Mg²⁺ + 2OH⁻
The ration of Mg(OH)₂ and OH⁻ is 1:2.
[OH⁻] = 2× 0.100 = 0.200 M
C =0.100 M ammonia
NH₃ + H₂O → NH₄OH → NH₄⁺ + OH⁻
The Kb expression will be written as,
Kb = [NH₄⁺] [OH⁻] / [NH₃]
Kb = 1.8 ×10 ⁻⁵
Now we will put the values:
1.8 ×10 ⁻⁵ = [NH₄⁺] [OH⁻] / 0.100
[NH₄⁺] [OH⁻] = x²
1.8 ×10 ⁻⁵ = x² / 0.100
x² = 1.8 ×10 ⁻⁵ × 0.100
x² = 1.8 × 10 ⁻⁶
x = 1.34 × 10 ⁻³
[OH⁻] = 1.34 × 10 ⁻³ M
d = 0.300 M rubidium hydroxide
RbOH + H₂O → Rb⁺ + OH⁻
The ratio of RbOH and OH⁻ is 1:1.
[OH⁻] = 0.300 M
e = 0.250 M calcium hydroxide
Ca(OH)₂ + H₂O → Ca²⁺ + 2OH⁻
The ration of Ca(OH)₂ and OH⁻ is 1:2.
[OH⁻] = 2× 0.250 = 0.5 M
