# A 5.00*mL volume of acetic acid required a 45*mL volume of 0.100*mol*L^-1 NaOH for equivalence. What is the concentration of the acetic acid solution?

Nov 20, 2016

Approx. $1 \cdot m o l \cdot {L}^{-} 1$, with respect to acetic acid.

#### Explanation:

$\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$, or $C = \frac{n}{V}$, and clearly $\text{concentration}$ has units of $m o l \cdot {L}^{-} 1$.

We need a reaction:

H_3C-CO_2H(aq) + NaOH(aq) rarr H_3C-CO_2^(-)""^(+)Na(aq) + H_2O(aq)

And thus, since
$\text{moles of sodium hydroxide "-=" moles of acetic acid}$,

there were $45.0 \cdot m L \times {10}^{-} 3 L \cdot m {L}^{-} 1 \times 0.100 \cdot m o l \cdot {L}^{-} 1 =$ $4.50 \times {10}^{-} 3 \cdot m o l$ of acetic acid in that original $5 \cdot m L$ volume.

Again, $\text{Concentration}$ $=$ $\text{Moles of solute"/"Volume of solution}$

$=$ $\frac{4.50 \times {10}^{-} 3 \cdot m o l}{5 \times {10}^{-} 3 L}$