# Question #72f34

Dec 10, 2017

$\left[H C l\right] \cong 0.065 \cdot m o l \cdot {L}^{-} 1$

#### Explanation:

We first write the stoichiometric equation....

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

...to establish the 1:1 equivalence...

And thus $\left[H C l\right] = \frac{25.81 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1 \times 0.1250 \cdot m o l \cdot {L}^{-} 1}{50.00 \times {10}^{-} 3 \cdot L}$

$= 0.06453 \cdot m o l \cdot {L}^{-} 1$

Dec 10, 2017

$\text{0.065 M}$

#### Explanation:

The formula for the reaction is

$\text{HCl" + "KOH" -> "KCl" + "H"_2"O}$

and its already balanced (yea!).

When the mL and M of something $\left(\text{KOH}\right)$ is givenm use cross multiplication to find how many mols (remember $\text{M}$ is $\text{mol/liter}$, or $\text{mol/1000 mL}$):

$\text{0.1250 mol"/"1000 mL" = "x mol"/ "25.81 mL}$

to get $3.23 \times {10}^{-} 3$ $\text{moles KOH}$, and since in the reaction equation all coefficients are 1, that means there were $3.23 \times {10}^{-} 3$ $\text{moles}$ of $\text{HCl}$ in the $\text{50 mL}$ sample.

Then, you use cross multiplication again to find how many moles would be in $\text{1000 mL}$ of that sample:

$\text{x mol"/"1000 mL" = (3.23xx10^-3 "mol")/ "50 mL}$

which will give you $\text{0.065 mol/1000 mL}$, which is $\text{0.065 M}$.