What is $[HCl]$ if a $17.04mL$ volume of $0.468molL^-1$ concentration of $[Ca(OH)_2]$ was titrated with $13.7mL$ of said acid?

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Answer 1 · 2016-05-25T16:03:43

Approx. $1* mol*L^-1$

We need (i) the balanced chemical equation:

$2HCl(aq) + Ca(OH)_2(aq) rarr CaCl_2(aq) + 2H_2O(l)$

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

$"Moles of calcium hydroxide"$ $=$ $17.04xx10^-3Lxx0.468*mol*L^-1$ $=$ $7.98xx10^-3*mol$

Twice this molar quantity of $HCl$ was present in the $13.7*mL$ $HCl(aq)$ .

Thus $[HCl]=(2xx7.98xx10^-3*mol)/(13.7xx10^-3*L)$ $=$ $??mol*L^-1$

Note that the concentration of $Ca(OH)_2$ is unreasonable....

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?