# What volume in milliliters of 9.950*10^(−2)"M" sodium hydroxide solution is required to reach the equivalence point in the complete titration of a 17.0mL sample of 0.114M phosphoric acid?

May 26, 2015

You'd need 58.4 mL of sodium hydroxide solution to neutralize that much phosphoric acid.

Start by writing the balanced chemical equation for your neutralization reaction

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

Notice that you you have a $1 : \textcolor{red}{3}$ mole ratio between phosphoric acid and sodium hydroxide. This tells you that, regardless of how many moles of phosphoric acid are present in the initial solution, you'll always need 3 times more moles of sodium hydroxide for the reaction to take place.

Use the phosphoric acid solution's molarity and volume to determine how many moles of acid you have

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

${n}_{{H}_{3} P {O}_{4}} = \text{0.114 M" * 17.0 * 10^(-3)"L" = "0.001938 moles}$

Use the mole ratio to figure out how many moles of sodium hydroxide you'd have to add to neutralize this much acid

0.001938cancel("moles"H_3PO_4) * (color(red)(3)"moles NaOH")/(1cancel("mole"H_3PO_4)) = "0.005814 moles NaOH"

Now use the molarity of the sodium hydroxide solution to see what volume would contain this many moles

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

V_(NaOH) = (0.005814cancel("moles"))/(9.950 * 10^(-2)cancel("moles")/"L") = "0.05843 L"

Expressed in mL and rounded to three sig figs, the number of sig figs you gave for the volume of the phosphoric acid solution, the answer will be

${V}_{N a O H} = \textcolor{g r e e n}{\text{58.4 mL}}$

SIDE NOTE To see what's going on at the ion level, you can remove the spectator ions, which are ions that are present on both sides of the equation, from the balanced chemical equation to get the net ionic equation

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