How to balance lead and silver acetate yields lead (||) acetate and silver
1 answer:
The given elements put into an equation using their symbols are as follows:
Pb +
= 
+ Ag
Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb + = + Ag
Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb +
= + Ag
Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb + = + 2Ag
That is your final equation
The coefficients are 2 + 2 = 1 + 2
Pb +
Since there are 2 Pb on the right side of the equation, you would change the coefficient of Pb on the left side to 2:
2Pb +
Since there are 2 Acetate on the right side of the equation, you would change the coefficient of Silver Acetate on the left side to 2:
2Pb +
Now there are 2 Silver on the left side, so you change the coefficient of Silver on the right side to 2:
2Pb +
That is your final equation
The coefficients are 2 + 2 = 1 + 2
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Answer:
The [SO₃²⁻]
Explanation:
From the first dissociation of sulfurous acid we have:
H₂SO₃(aq) ⇄ H⁺(aq) + HSO₃⁻(aq)
At equilibrium: 0.50M - x x x
The equilibrium constant (Ka₁) is:
With Ka₁= 1.5x10⁻² and solving the quadratic equation, we get the following HSO₃⁻ and H⁺ concentrations:
Similarly, from the second dissociation of sulfurous acid we have:
HSO₃⁻(aq) ⇄ H⁺(aq) + SO₃²⁻(aq)
At equilibrium: 7.94x10⁻²M - x x x
The equilibrium constant (Ka₂) is:
Using Ka₂= 6.3x10⁻⁸ and solving the quadratic equation, we get the following SO₃⁻ and H⁺ concentrations:
Therefore, the final concentrations are:
[H₂SO₃] = 0.5M - 7.94x10⁻²M = 0.42M
[HSO₃⁻] = 7.94x10⁻²M - 7.07x10⁻⁵M = 7.93x10⁻²M
[SO₃²⁻] = 7.07x10⁻⁵M
[H⁺] = 7.94x10⁻²M + 7.07x10⁻⁵M = 7.95x10⁻²M
So, the lowest concentration at equilibrium is [SO₃²⁻] = 7.07x10⁻⁵M.
I hope it helps you!