An unknown solution has a ph of 8. how would you classify this solution?
1 answer:
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The number of liters of 3.00 M lead (II) iodide : 0.277 L
<h3>Further explanation</h3>Reaction(balanced)
Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)
moles of KI = 1.66
From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :
Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution
Where
M = Molarity
n = Number of moles of solute
V = Volume of solution
So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):
Answer:
Molar concentration of S₂ is 1.77×10⁻⁶M
Explanation:
For the reaction:
2H₂S(g) ⇄ 2H₂(g) + S₂(g)
The equilibirum constant, K, is defined as:
<em>(1)</em>
Concentrations in equilibirum are:
[H₂S] : 0,163/0.500L - X
[H₂] : 0,0500/0.500L + X
[S₂] : X
Replacing the concentrations and the equilibrium value in (1):
1.67x10⁻⁷ = X (X² + 0.2X + 0.01) / (X² -0.652X + 0.106)
1.67x10⁻⁷X² - 1.09x10⁻⁷X + 1.77x10⁻⁸ = X³ + 0.2X² + 0.01X
0 = X³ + 0.2X² + 0.01X - 1.77x10⁻⁸
Solving for X:
X = 1.77×10⁻⁶
As [S₂] = X, <em>molar concentration of S₂ is 1.77×10⁻⁶M</em>
I hope it helps!