# Question 178d9

Apr 28, 2015

You'd need 0.24 L of phosphoric acid solution to react with that much sodium hydroxide.

${H}_{3} P {O}_{4 \left(a q\right)} + \textcolor{red}{3} N a O {H}_{\left(a q\right)} \to N {a}_{3} P {O}_{4 \left(a q\right)} + 3 {H}_{2} {O}_{\left(l\right)}$

Notice that you have a $1 : \textcolor{red}{3}$ mole ratio between phosporic acid and sodium hydroxide. This tells you that, regardless of how many moles of the former react, you need 3 times more moles of sodium hydroxide to completely neutralize the acid.

Since you know the molarities of the solutions, and the volume of the sodium hydroxide solution, you can determine how many moles of sodium hydroxide are added

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

${n}_{N a O H} = C \cdot V = \text{3.8 M" * "0.347 L" = "1.3186 moles NaOH}$

You can now use the aforementioned mole ratio to see how many moles of acid would be neutralized by this many moles of sodium hydroxide

1.3186cancel("moles NaOH") * ("1 mole "H_3PO_4)/(color(red)(3)cancel("moles NaOH")) = "0.4395 moles"#

Now use the solution's molarity to determine the volume

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

${V}_{{H}_{3} P {O}_{4}} = \text{0.4395 moles"/"1.8 M" = "0.2442 L}$

Rounded to two sig figs, the number of sig figs given for 1.8 M and 3.8 M, the answer will be

${V}_{{H}_{3} P {O}_{4}} = \textcolor{g r e e n}{\text{0.24 L}}$