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DIA [1.3K]
3 years ago
15

What is the solubility in pure water of ba(io3)2 in moles per liter at 25 ˚c? [ksp (25 ˚c) = 6.0  10–10]?

Chemistry
1 answer:
diamong [38]3 years ago
6 0
Ba(IO₃)₂(s) partially dissociates in water  as Ba²⁺(aq) and IO₃⁻(aq).
                Ba(IO₃)₂(s) ⇄ Ba²⁺(aq) + 2IO₃⁻(aq)
Initial          
Change        -X                +X             2X
Equilibrium                        X              2X

           Ksp = [Ba²⁺(aq)] x [IO₃⁻(aq)]²
6.0 x 10⁻¹⁰  = X * (2X)²
6.0 x 10⁻¹⁰ = 4X³
              X = 5.313 x 10⁻⁴ mol/L

Hence, the solubility of the Ba(IO₃)₂(s) is 5.313 x 10⁻⁴ mol/L

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Explanation:

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This is a mathematical question, instead of a chemistry question, and you should use calculus to find the nitrogen level that gives the best yield, since this is an optimization problem.

The best yield is the maximum yield, and the maximum, provided that it exists, is found using the first derivative and making it equal to zero: Y' = 0

To find Y' you must use the quotient rule.

Y'=\frac{(kN)'(9+N^2)- (kN)(9+N^2)'}{(9+N^2)^2}\\ \\Y'=\frac{k(9+N^2)-kN(2N)}{(9+N^2)^2}\\ \\Y'=\frac{9k + kN^2 - 2kN^2}{(9 + N^2)^2}=\frac{9k-kN^2}{(9+N^2)^2}

Now make Y' = 0

  • The denominator is never equal to zero, because it is always positive and greater than 9.

  • Make the numerator equal to zero:

         9k - kN² = 0

  • Factor: k (9 - N²) = 0

  • Since k is a positve constant, it is not equal to zero, and the other factor, 9 - N², must be equal to zero:

         9 - N² = 0 ⇒ (3 - N) (3 + N) = 0

         ⇒ 3 - N = 0 or 3 + N = 0 ⇒ N = 3 or N = -3.

Since N is nitrogen level, it cannot be negative and the only valid answer is N = 3.

You can prove that it is a maximum (instead of a minimum) finding the second derivative or testing some points around 3 (e.g. 2.5 and 3.5).

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