A 13.8 mL aliquot of 0.176 M H₃PO₄(aq) is to be titrated with 0.110 M NaOH (aq). What volume (mL) of base will it take to reach the equivalence point? A) 20.9 B) 7.29 C) 199 D) 66.2 E) 22.1

Question

Quizplus · Accepted Answer

We can use the balanced chemical equation for the reaction between HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0 and NaOH to determine the mole ratio between them.

HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0(aq) + NaOH(aq) → NaHS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1(aq) + H2O(l)

From the balanced equation, we see that one mole of NaOH reacts with one mole of HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0. Therefore, the number of moles of NaOH required to react completely with HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0 is:

moles of NaOH = Molarity × Volume (L) = 0.110 mol/L × 13.8 mL ÷ 1000 mL/L = 0.001518 mol

To reach the equivalence point, the number of moles of NaOH must be equal to the number of moles of HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0:

moles of HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0 = Molarity × Volume (L) = 0.176 mol/L × 0.0138 L = 0.0024288 mol

Now we can use the mole ratio to determine the volume of NaOH required:

moles of NaOH ÷ moles of HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0 = Volume (L) of NaOH ÷ Volume (L) of HS1U1B13S1U1B0POS1U1B14S1U1B0S1U1B1S1U1B0

0.001518 mol ÷ 0.0024288 mol = Volume (L) of NaOH ÷ 0.0138 L

Volume (L) of NaOH = 0.001518 mol × 0.0138 L ÷ 0.0024288 mol = 0.00862 L = 8.62 mL

Therefore, the answer is D) 66.2 mL.

A 138 ML Aliquot of 0