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Orlov [11]
3 years ago
15

Determine the molar solubility of pbso4 in pure water. ksp (pbso4) = 1.82 x 10-8.

Chemistry
1 answer:
Misha Larkins [42]3 years ago
5 0
Use the ICE table approach as solution:

           PbSO₄   --> Pb²⁺ + SO₄²⁻
I             -                 0          0
C           -                +s         +s
E           -                  s          s

Ksp = [Pb²⁺][SO₄²⁻]
1.82×10⁻⁸ = s²
Solving for s,
s = <em>1.35×10⁻⁴ M</em>
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3 years ago
At 25°C, the equilibrium constant Kc for the reaction 2A(aq) ↔ B(aq) + C(aq) is 65. If 2.50 mol of A is added to enough water to
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Answer:

0.146 M

Explanation:

Equation for the reaction :

2A(aq) ↔ B(aq) + C(aq)

K_c = 65

Molar concentration of A = \frac{2.50 mol}{1.00 L}

= 2.5 M

                       2A(aq)     ↔        B(aq)     +   C(aq)

Initial              2.50                      0                0

Change          - 2x                      + x             + x

Equilibrium   2.50 - 2x               +x               +x

K_c =\frac {[B][C]}{[A]^2}

65 = \frac{[x][x]}{[2.5-2x]^2}

65 = \frac{[x]^2}{[2.5-2x]^2}

65 = (\frac{[x]}{[2.5-2x]})^2

\sqrt 65 =  \sqrt {(\frac{[x]}{[2.5-2x]})^2}

8.062 =  \frac{x}{2.5-2x}

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20.155 - 16.124x = x

20.155 = 16.124x+x

20.155 = 17.124x

x = \frac{20.155}{17.124}

x = 1.177

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