$2 \times 10^{- 3} \cdot L$ $2xx10^-3L$ of $N a O H$ $NaOH$ of $0.200 \cdot m o l \cdot L^{- 1}$ $0.200molL^-1$ concentration is added to $1.00 \cdot L$ $1.00L$ of $1.00 \cdot m o l \cdot L^{- 1}$ $1.00molL^-1$ $C a {(N O_{3})}_{2} (a q)$ $Ca(NO_3)_2(aq)$ ...will calcium hydroxide precipitate?

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$2 \times 10^{- 3} \cdot L$ $2xx10^-3L$ of $N a O H$ $NaOH$ of $0.200 \cdot m o l \cdot L^{- 1}$ $0.200molL^-1$ concentration is added to $1.00 \cdot L$ $1.00L$ of $1.00 \cdot m o l \cdot L^{- 1}$ $1.00molL^-1$ $C a {(N O_{3})}_{2} (a q)$ $Ca(NO_3)_2(aq)$ ...will calcium hydroxide precipitate?

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Answer 1 · 2017-03-10T08:47:31

Calcium hydroxide has a solubility of $1.73 \cdot g \cdot L^{- 1}$ $1.73*g*L^-1$ under standard conditions. This should have been quoted in the question. Calcium hydroxide should not precipitate.

Explanation:

We work out the equivalent quantity of calcium hydroxide, for which we (i) need a stoichiometric equation:

$C a^{2 +} + 2 H O^{-} \to C a {(O H)}_{2} (s) ⏐ ⏐ ↓$ $Ca^(2+) + 2HO^(-) rarr Ca(OH)_2(s) darr$

And (ii) equivalent quantities of the ions in solution:

$Moles of sodium hydroxide,$ $"Moles of sodium hydroxide,"$ $=$ $=$ $2.00 \times 10^{- 3} L \times 0.200 \cdot m o l \cdot L^{- 1} = 4.0 \times 10^{- 4} \cdot m o l$ $2.00xx10^-3Lxx0.200*mol*L^-1=4.0xx10^-4*mol$ .

And thus when this is added to the bulk solution, we have $[H O^{-}]$ $[HO^-]$ $=$ $=$ $\frac{4.0 \times 10^{- 4} \cdot m o l}{1.002 \cdot L} = 4.0 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $(4.0xx10^-4*mol)/(1.002*L)=4.0xx10^-4*mol*L^-1$ .

Given the stoichiometry, we thus have a solution that is NOMINALLY $2.0 \times 10^{- 4} \cdot m o l \cdot L^{- 1}$ $2.0xx10^-4*mol*L^-1$ with respect to calcium hydroxide.

And this is a mass solubility of,

$74.09 \cdot g \cdot m o l^{- 1} \times 2.0 \times 10^{- 4} \cdot m o l \cdot L^{- 1} = 14.8 \cdot m g \cdot L^{- 1}$ $74.09*g*mol^-1xx2.0xx10^-4*mol*L^-1=14.8*mg*L^-1$ or $15 ppm$ $"15 ppm"$ .

And so calcium hydroxide should NOT precipitate.

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$2 \times 10^{- 3} \cdot L$ $2xx10^-3L$ of $N a O H$ $NaOH$ of $0.200 \cdot m o l \cdot L^{- 1}$ $0.200molL^-1$ concentration is added to $1.00 \cdot L$ $1.00L$ of $1.00 \cdot m o l \cdot L^{- 1}$ $1.00molL^-1$ $C a {(N O_{3})}_{2} (a q)$ $Ca(NO_3)_2(aq)$ ...will calcium hydroxide precipitate?

1 Answer

Explanation:

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2×10−3⋅L2xx10^-3*L of NaOHNaOH of 0.200⋅mol⋅L−10.200*mol*L^-1 concentration is added to 1.00⋅L1.00*L of 1.00⋅mol⋅L−11.00*mol*L^-1 Ca(NO3)2(aq)Ca(NO_3)_2(aq)...will calcium hydroxide precipitate?

1 Answer

Explanation:

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$2 \times 10^{- 3} \cdot L$ $2xx10^-3L$ of $N a O H$ $NaOH$ of $0.200 \cdot m o l \cdot L^{- 1}$ $0.200molL^-1$ concentration is added to $1.00 \cdot L$ $1.00L$ of $1.00 \cdot m o l \cdot L^{- 1}$ $1.00molL^-1$ $C a {(N O_{3})}_{2} (a q)$ $Ca(NO_3)_2(aq)$ ...will calcium hydroxide precipitate?