Given:
35.0 mL of acid with an unknown concentration
24.6 mL of 0.432 M base
Required:
Concentration of the acid
Solution:
M1V1 = M2V2
M1 (35.0 mL of acid)
= (0.432 M base) (24.6 mL
of base)
V1 = (0.432
M base) (24.6 mL of base) /
(35.0 mL of acid)
M1 = 0.304 M of acid
X is always the independent variable
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
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