What is the ionic equation for potassium fluoride and ammonium nitrate
1 answer:
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Answer:
the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷
Explanation:
Given the data in the question;
+
⇄
Formation constant Kf
Kf = / ( [
][
] ) = 5.0 × 10¹⁰
Now,
[] =
; ∝₄ = 0.35
so the equilibrium is;
+
⇄
+ 4H⁺
Given that; =
{ 1 mol
reacts with 1 mol
}
so at equilibrium, =
= x
∴
+
⇄
x + x 0.010-x
since Kf is high, them x will be small so, 0.010-x is approximately 0.010
so;
Kf = / ( [
][
] ) =
/ ( [
][
] ) = 5.0 × 10¹⁰
⇒ / ( [
][
] ) = 5.0 × 10¹⁰
⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰
⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰
⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 0.010 / 1.75 × 10¹⁰
⇒ x² = 5.7142857 × 10⁻¹³
⇒ x = √(5.7142857 × 10⁻¹³)
⇒ x = 7.559 × 10⁻⁷
Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷