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What is the volume of 0.100 M sodium hydroxide needed to precipitate all of the nickel(II) ions from 150.0 mL of 0.250 M nickel(II) nitrate?

Aug 21, 2016

We need a volume of $0.75 \cdot L$

Explanation:

$N {i}^{2 +} + 2 O {H}^{-} \rightarrow N i {\left(O H\right)}_{2} \left(s\right) \downarrow$

Alternatively,

$N i {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 N a O H \left(a q\right) \rightarrow N i {\left(O H\right)}_{2} \left(s\right) + 2 N a N {O}_{3} \left(a q\right)$

We need a stoichiometrically balanced equation to inform our calculations.

$\text{Moles of nickel ion}$ $=$ $150.0 \times {10}^{-} 3 L \times 0.250 \cdot m o l \cdot {L}^{-} 1 = 0.0375 \cdot m o l$.

Because of the stoichiometry, we need $2 \times 0.0375 \cdot m o l \text{ hydroxide ion} = 0.0750 \cdot m o l$.

And thus volume of $0.100 \cdot m o l \cdot {L}^{-} 1$ $N a O H$ $=$ $\frac{0.0750 \cdot m o l}{0.100 \cdot m o l \cdot {L}^{-} 1} \times {10}^{3} \cdot m L \cdot {L}^{-} 1$ $=$ $750 \cdot m L$

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