Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
Answer:
8
Explanation:
1 mole = 6.02 × 10²³ atoms
? moles = 4.816 × 10²⁴ atoms.
? Moles = 4.816 × 10²⁴ ÷ 6.02 × 10²³
? Moles = 8 moles
8 moles of aluminum = 4.816 × 10²⁴
Answer : The volume of stock solution needed are, 12.5 mL
Explanation :
Formula used :
where,
are the initial molarity and volume of copper (II) chloride.
are the final molarity and volume of stock solution of copper (II) chloride.
We are given:
Putting values in above equation, we get:
Hence, the volume of stock solution needed are, 12.5 mL
