What is $[H C l]$ $[HCl]$ if a $17.04 \cdot m L$ $17.04mL$ volume of $0.468 \cdot m o l \cdot L^{- 1}$ $0.468molL^-1$ concentration of $[C a {(O H)}_{2}]$ $[Ca(OH)_2]$ was titrated with $13.7 \cdot m L$ $13.7mL$ of said acid?

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What is $[H C l]$ $[HCl]$ if a $17.04 \cdot m L$ $17.04mL$ volume of $0.468 \cdot m o l \cdot L^{- 1}$ $0.468molL^-1$ concentration of $[C a {(O H)}_{2}]$ $[Ca(OH)_2]$ was titrated with $13.7 \cdot m L$ $13.7mL$ of said acid?

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Answer 1 · 2016-05-25T16:03:43

Approx. $1 \cdot m o l \cdot L^{- 1}$ $1* mol*L^-1$

Explanation:

We need (i) the balanced chemical equation:

$2 H C l (a q) + C a {(O H)}_{2} (a q) \to C a C l_{2} (a q) + 2 H_{2} O (l)$ $2HCl(aq) + Ca(OH)_2(aq) rarr CaCl_2(aq) + 2H_2O(l)$

And (ii) the amount in moles of calcium hydroxide.

$Moles of calcium hydroxide$ $"Moles of calcium hydroxide"$ $=$ $=$ $17.04 \times 10^{- 3} L \times 0.468 \cdot m o l \cdot L^{- 1}$ $17.04xx10^-3Lxx0.468*mol*L^-1$ $=$ $=$ $7.98 \times 10^{- 3} \cdot m o l$ $7.98xx10^-3*mol$

Twice this molar quantity of $H C l$ $HCl$ was present in the $13.7 \cdot m L$ $13.7*mL$ $H C l (a q)$ $HCl(aq)$ .

Thus $[H C l] = \frac{2 \times 7.98 \times 10^{- 3} \cdot m o l}{13.7 \times 10^{- 3} \cdot L}$ $[HCl]=(2xx7.98xx10^-3*mol)/(13.7xx10^-3*L)$ $=$ $=$ $? ? m o l \cdot L^{- 1}$ $??mol*L^-1$

Note that the concentration of $C a {(O H)}_{2}$ $Ca(OH)_2$ is unreasonable....

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What is $[H C l]$ $[HCl]$ if a $17.04 \cdot m L$ $17.04mL$ volume of $0.468 \cdot m o l \cdot L^{- 1}$ $0.468molL^-1$ concentration of $[C a {(O H)}_{2}]$ $[Ca(OH)_2]$ was titrated with $13.7 \cdot m L$ $13.7mL$ of said acid?

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What is [HCl][HCl] if a 17.04⋅mL17.04*mL volume of 0.468⋅mol⋅L−10.468*mol*L^-1 concentration of [Ca(OH)2][Ca(OH)_2] was titrated with 13.7⋅mL13.7*mL of said acid?

1 Answer

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What is $[H C l]$ $[HCl]$ if a $17.04 \cdot m L$ $17.04mL$ volume of $0.468 \cdot m o l \cdot L^{- 1}$ $0.468molL^-1$ concentration of $[C a {(O H)}_{2}]$ $[Ca(OH)_2]$ was titrated with $13.7 \cdot m L$ $13.7mL$ of said acid?