Why does a $23 \cdot g$ $23g$ mass of calcium hydroxide dissolved in a $400 \cdot m L$ $400mL$ volume of water, have $[H O^{-}] \equiv 1.55 \cdot m o l \cdot L^{- 1}$ $[HO^-]-=1.55molL^-1$ ?

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Why does a $23 \cdot g$ $23g$ mass of calcium hydroxide dissolved in a $400 \cdot m L$ $400mL$ volume of water, have $[H O^{-}] \equiv 1.55 \cdot m o l \cdot L^{- 1}$ $[HO^-]-=1.55molL^-1$ ?

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Answer 1 · 2016-10-16T18:48:50

It seems that you have grasped the concept of stoichiometry:

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

Explanation:

$[H O^{-}]$ $[HO^-]$ $=$ $=$ $\frac{Moles of hydroxide ion}{Volume of solution}$ $"Moles of hydroxide ion"/"Volume of solution"$

$Moles of calcium hydroxide = \frac{23 \cdot g}{74.1 \cdot g \cdot m o l^{- 1}}$ $"Moles of calcium hydroxide"=(23*g)/(74.1*g*mol^-1)$ $=$ $=$ $0.310 \cdot m o l$ $0.310*mol$ . $Moles of calcium = 0.155 \cdot m o l$ $"Moles of calcium"=0.155*mol$ .

$[H O^{-}]$ $[HO^-]$ $=$ $=$ $\frac{2 \times 0.310 \cdot m o l}{0.400 \cdot L}$ $(2xx0.310*mol)/(0.400*L)$ $=$ $=$ $1.55 \cdot m o l \cdot L^{- 1}$ $1.55*mol*L^-1$ , because each formula unit of $C a {(O H)}_{2}$ $Ca(OH)_2$ delivers one $C a^{2 +}$ $Ca^(2+)$ , but $2 \times H O^{-}$ $2xxHO^-$ .

Each mole of calcium hydroxide clearly gives 2 moles of hydroxide anion upon dissolution, simply because of the formulation of calcium hydroxide. Is this clear?

To put it another way, the given solution is $1.55 \cdot m o l \cdot L^{- 1}$ $1.55*mol*L^-1$ with respect to $H O^{-}$ $HO^-$ , BUT half this concentration with respect to $C a {(O H)}_{2} (a q)$ $Ca(OH)_2(aq)$ and $C a^{2 +}$ $Ca^(2+)$ .

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Why does a $23 \cdot g$ $23g$ mass of calcium hydroxide dissolved in a $400 \cdot m L$ $400mL$ volume of water, have $[H O^{-}] \equiv 1.55 \cdot m o l \cdot L^{- 1}$ $[HO^-]-=1.55molL^-1$ ?

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Why does a 23⋅g23*g mass of calcium hydroxide dissolved in a 400⋅mL400*mL volume of water, have [HO−]≡1.55⋅mol⋅L−1[HO^-]-=1.55*mol*L^-1?

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Explanation:

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Why does a $23 \cdot g$ $23g$ mass of calcium hydroxide dissolved in a $400 \cdot m L$ $400mL$ volume of water, have $[H O^{-}] \equiv 1.55 \cdot m o l \cdot L^{- 1}$ $[HO^-]-=1.55molL^-1$ ?