# If 36.10 mL of 0.223 M of NaOH is used to neutralize a 0.515 g sample of citric acid, what is the molar mass of the acid?

Apr 6, 2015

The molar mass of the citric acid is $\text{192 g/mol}$.

The balanced chemical equation for the neutralization reaction that takes place between sodium hydroxide, $N a O H$, and citric acid, ${C}_{6} {H}_{8} {O}_{7}$, looks like this

${C}_{6} {H}_{8} {O}_{7 \left(a q\right)} + \textcolor{red}{3} N a O {H}_{\left(a q\right)} \to N {a}_{3} {C}_{6} {H}_{5} {O}_{7 \left(a q\right)} + 3 {H}_{2} {O}_{\left(l\right)}$

Notice the $1 : \textcolor{red}{3}$ mole ratio that exists between citric acid and sodium hydroxide; what this tells you is that, for every mole of citric acid, you need 3 times more moles of sodium hydroxide for the reaction to take place.

Since you know the molarity and the volume of the $N a O H$ you've used, you can calculate how many moles of $N a O H$ reacted

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{N a O H} = \text{0.223 M" * 36.10 * 10^(-3)"L" = "0.00805 moles}$

Now use the aforementioned mole ratio to see how many moles of citric acid were present in 0.515 g

0.00805cancel("moles NaOH") * "1 mole citric acid"/(3cancel("moles NaOH")) = "0.00268 moles citric acid"

Now simply divide the mass of citric acid given by the number of moles it contained to get the compound's molar mass

${M}_{M} = \frac{m}{n} = \text{0.515 g"/("0.00268 moles") = "192.16 g/mol}$

Rounded to three sig figs, the number of sig figs given for 0.515 g, the answer will be

${M}_{M} = \textcolor{g r e e n}{\text{192 g/mol}}$

SIDE NOTE The actual molar mass of citric acid is 192.12 g/mol, so your result is in agreement with the known value.