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3 years ago

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To what final concentration of NH3 must a solution be adjusted to just dissolve 0.060 mol of NiC2O4 (Ksp = 4×10−10) in 1.0 L of

solution? The Kf for Ni(NH3)62+ = 1.2 x 109

1 answer:

ololo11 [35]3 years ago

3 0

Given:
0.060 mol of NiC2O4
Ksp = 4 x 10⁻¹⁰
1.0 L of solution
Kf of Ni(NH3)6 2⁺ = 1.2 x 10⁹
<span>NiC2O4 + 6NH3 ⇋ Ni(NH3)6 2+ + 2O4 2- </span>
<span>NiC2O4 ⇋ Ni 2+ + C2O4 2- ...Ksp </span>
<span>Ni2+ + 6NH3 ⇋ Ni(NH3)6 2+...Kf </span>

Ksp * Kf = (4 x 10⁻¹⁰) * (1.2 x 10⁹) = 0.48

K = 0.48 = [Ni(NH3)6 2+][C2O4 2-] / [NH3]⁶<span>
</span>0.48 = (0.060)² / [NH3]⁶<span> ... (dissolved C2O4 2- = 0.060M)
</span><span>[NH3]</span>⁶<span> = (0.060)</span>²<span> / 0.48 = </span>0.0036 / 0.48 = 0.0075
NH3 = ⁶√0.0075

NH3 = 0.44 M

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