# Question 536b8

Mar 13, 2015

The molarity of the sodium hydroxide solution is $\text{0.1632 M}$.

The balanced chemical equation for this titration, in which you're essentially performing a neutralization reaction, looks like this

${H}_{2} {C}_{2} {O}_{4 \left(A q\right)} + 2 N a O {H}_{\left(a q\right)} \to N {a}_{2} {C}_{2} {O}_{4 \left(a q\right)} + 2 {H}_{2} {O}_{\left(l\right)}$

In order for the oxalic acid to solution to completely neutralize the sodium hydroxide solution, you must add exactly 1 mole of oxalic acid for every 2 moles of sodium hydroxide.

This results from the $\text{1:2}$ mole ratio that exists between the two compounds - 1 mole of oxalic acid needs 2 moles of sodium hydroxide for the reaction to takes place.

The number of moles of oxalic acid used can be calculated using the solution's molarity

C = n/V => n_("oxalic acid") = C * V = 20.34 * 10^(-3)"L" * "0.1003 M"

n_("oxalic acid") = = "0.002040 moles oxalic acid"

According to the aforementioned mole ratio, the number of sodium hydroxide moles must be

$\text{0.002040 moles oxalic acid" * "2 moles sodium hydroxide"/"1 mole oxalic acid" = "0.004080 moles sodium hydroxide}$

FInally, use the volume of the sodium hydroxide solution to solve for its concentration

C = n_("sodium hydroxide")/V_("sodium hydroxide") = "0.004080 moles"/(25 * 10^(-3)"L") = "0.1632 M"#