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raketka [301]
3 years ago
11

The solubility of CaF2 is 0.00021 mole per liter. What is the solubility product constant for CaF2? A) 7.3 x 10-12 B) 3.7 x 10-1

1 C) 8.5 x 10-8 D) 4.4 x 10-8 E) 1.9 x 10-11
Chemistry
1 answer:
Naddika [18.5K]3 years ago
3 0

Answer:

B) Ksp = 3.7 E-11

Explanation:

CaF2 ↔ Ca2+  +  2F-

  S            S          2S,,,,,,,,,,,,,,,in the equilibrium

⇒ Ksp = [ Ca2+ ] * [ F- ]²

⇒ Ksp = S * ( 2S )²

⇒ Ksp = 4S³

⇒ Ksp = 4 * ( 0.00021 )³

⇒ Ksp = 4 * 9.261 E-12

⇒ Ksp = 3.704 E-11

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{\displaystyle K_{\rm {eq}}={\frac {a_{\rm {H_{3}O^{+}}}\cdot a_{\rm {OH^{-}}}}{a_{\rm {H_{2}O}}^{2}}}}

which is numerically equal to the more traditional thermodynamic equilibrium constant written as:

{\displaystyle K_{\rm {eq}}={\frac {a_{\rm {H^{+}}}\cdot a_{\rm {OH^{-}}}}{a_{\rm {H_{2}O}}}}}

under the assumption that the sum of the chemical potentials of H+ and H3O+ is formally equal to twice the chemical potential of H2O at the same temperature and pressure.[1]

Because most acid–base solutions are typically very dilute, the activity of water is generally approximated as being equal to unity, which allows the ionic product of water to be expressed as:[2]

{\displaystyle K_{\rm {eq}}\approx a_{\rm {H_{3}O^{+}}}\cdot a_{\rm {OH^{-}}}}

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{\displaystyle K_{\rm {w}}=[{\rm {H_{3}O^{+}}}][{\rm {OH^{-}}}]}

where [H3O+] is the molarity (≈ molar concentration)[3] of hydrogen or hydronium ion, and [OH−] is the concentration of hydroxide ion. When the equilibrium constant is written as a product of concentrations (as opposed to activities) it is necessary to make corrections to the value of {\displaystyle K_{\rm {w}}} depending on ionic strength and other factors (see below).[4]

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