Answer:
[Cl⁻] = 0.016M
Explanation:
First of all, we determine the reaction:
Pb(NO₃)₂ (aq) + MgCl₂ (aq) → PbCl₂ (s) ↓ + Mg(NO₃)₂(aq)
This is a solubility equilibrium, where you have a precipitate formed, lead(II) chloride. This salt can be dissociated as:
PbCl₂(s) ⇄ Pb²⁺ (aq) + 2Cl⁻ (aq) Kps
Initial x
React s
Eq x - s s 2s
As this is an equilibrium, the Kps works as the constant (Solubility product):
Kps = s . (2s)²
Kps = 4s³ = 1.7ₓ10⁻⁵
4s³ = 1.7ₓ10⁻⁵
s = ∛(1.7ₓ10⁻⁵ . 1/4)
s = 0.016 M
Answer:
0.080 mol
Explanation:
M(HCl) = (1.0 +35.5) g/mol = 36.5 g/mol
2.9g*1mol/36.5 g = 0.0795 mol HCl
Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O
from reaction 2 mol 2 mol
given 0.0795 mol x mol
x = 0.0795 mol ≈0.080 mol
if you have 1.27*10^-36= [Cu2+][X2-] then you can set both those values equal to x because they're stoichimetrically equal.
1.27*10^-36=x^2
Take the square root of both sides.
1.13*10^-18=x
That would be your solubility.
