# A student neutralizes 20.0 mL of a sodium hydroxide solution (NaOH) by adding 28.0 mL at 1.0 M hydrochloric acid (HCI). What is the molarity of the solution?

Nov 23, 2015

$\implies {\left({C}_{M}\right)}_{N a O H} = 1.4 M$

#### Explanation:

The net ionic equation is:

${H}^{+} \left(a q\right) + O {H}^{-} \left(a q\right) \to {H}_{2} O \left(l\right)$

Therefore, n_(H^+)=n_(OH^(-)

The relationship between number of mole ($n$) and molarity $\left({C}_{M}\right)$ is: ${C}_{M} = \frac{n}{V}$, where $V$ is the volume of the solution.

Thus, $n = {C}_{M} \times V$.

Therefore, (C_MxxV)_(H^+)=(C_MxxV)_(OH^(-)

$\implies {\left({C}_{M}\right)}_{O {H}^{-}}$$= \frac{{\left({C}_{M} \times V\right)}_{{H}^{+}}}{V} = \frac{1.0 M \times 28.0 \cancel{m L}}{20.0 \cancel{m L}} = 1.4 M$

$\implies {\left({C}_{M}\right)}_{N a O H} = 1.4 M$

Here is a video that might help you with the experimental procedure and calculations for the titration.
Lab Demonstration | Acid - Base Titration.