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

May 25, 2016

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

#### Explanation:

We need (i) the balanced chemical equation:

$2 H C l \left(a q\right) + C a {\left(O H\right)}_{2} \left(a q\right) \rightarrow C a C {l}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$

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

$\text{Moles of calcium hydroxide}$ $=$ $17.04 \times {10}^{-} 3 L \times 0.468 \cdot m o l \cdot {L}^{-} 1$ $=$ $7.98 \times {10}^{-} 3 \cdot m o l$

Twice this molar quantity of $H C l$ was present in the $13.7 \cdot m L$ $H C l \left(a q\right)$.

Thus $\left[H C l\right] = \frac{2 \times 7.98 \times {10}^{-} 3 \cdot m o l}{13.7 \times {10}^{-} 3 \cdot L}$ $=$ ??mol*L^-1

Note that the concentration of $C a {\left(O H\right)}_{2}$ is unreasonable....