# How many milliliters of .015 M NaOH are needed to neutralize 50.0 mL of 0.010 M HNO_3 (aq)? What compounds are formed after the reaction is complete?

Mar 9, 2016

${V}_{N a O H} = 33 m L$

#### Explanation:

this is a neutralization reaction between a strong acid $H C l$ and a strong base $N a O H$. The net ionic equation is:

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

since the acid and base are both monoprotic, therefore,

${n}_{{H}^{+}} = {n}_{O {H}^{-}} \implies {n}_{H C l} = {n}_{N a O H}$

${C}_{M} = \frac{n}{V} \implies n = {C}_{M} \times V$

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

$\implies {V}_{N a O H} = \frac{{\left({C}_{M} \times V\right)}_{H C l}}{{\left({C}_{M}\right)}_{N a O H}} = \frac{0.010 \cancel{M} \times 50.0 m L}{0.015 \cancel{M}} = 33 m L$