Question

A 35.00-mL sample of an unknown H3PO4 solution is titrated with a 0.130 M NaOH solution. The equivalence point is reached when 26.33 mL of NaOH solution is added. What is the concentration of the unknown H3PO4 solution? The neutralization reaction is: H3PO4(aq)+3NaOH(aq)→3H2O(l)+Na3PO4(aq)

Answer 1

Answer:

The concentration of the unknown H3PO4 solution is 0.0326 M

Explanation:

Step 1: Data given

Volume of the unknown H3PO4 solution = 35.00 mL

Molarity of NaOH solution = 0.130 M

The equivalence point is reached when 26.33 mL of NaOH solution is added.

Step 2: The balanced equation

H3PO4(aq) + 3NaOH(aq) → 3H2O(l) + Na3PO4(aq)

For 1 mole H3PO4, we need 3 moles NaOH to produce 3 moles of H2O and 1 mol of Na3PO4

Step 3: Calculate moles of NaOH

Moles NaOH = Molarity NaOH * volume

Moles NaOH = 0.130 M * 0.02633 = 0.0034229 moles NaOH

Step 4: Calculate moles of H3PO4

For 1 mole H3PO4, we need 3 moles NaOH to produce 3 moles of H2O and 1 mol of Na3PO4

 

For 0.0034229 moles NaOH, we have 0.0034229/3 = 0.00114 moles H3PO4 to react fully with the NaOH in the titrant.

 

Step 5: Calculate molarity of H3PO4

Molarity H3PO4 = moles H3PO4 / volume

Molarity H3PO4 = 0.00114 moles / 0.035 L

Molarity H3PO4 = 0.0326 M

The concentration of the unknown H3PO4 solution is 0.0326 M

Answer 2

The concentration of the unknown H₃PO₄ is 0.0326 M

From the question,

We are to determine the concentration of the unknown H₃PO₄

The given balanced chemical equation for the reaction is

H₃PO₄(aq) + 3NaOH(aq) → 3H₂O(l) + Na₃PO4(aq)

This means 1 mole of H₃PO₄ is needed to neutralize 3 moles of NaOH

Using the formula

$\frac{C_{A}V_{A} }{C_{B}V_{B}}= \frac{n_{A}}{n_{B}}$

Where $C_{A}$ is the concentration of acid

$C_{B}$ is the concentration of base

$V_{A}$ is the volume of acid

$V_{B}$ is the volume of base

$n_{A}$ is the mole ratio of acid

$n_{B}$ is the mole ratio of base

From the question

$V_{A} = 35.00 \ mL$

$C_{B} = 0.130 \ M$

$V_{B} = 26.33 \ mL$

From the balanced chemical equation

$n_{A} = 1$

$n_{B} = 3$

Putting the parameters into the formula, we get

$\frac{C_{A} \times 35.00}{0.130 \times 26.33} = \frac{1}{3}$

This becomes

$3 \times C_{A}\times 35.00 = 0.130 \times 26.33$

∴ $C_{A} =\frac{0.130 \times 26.33}{3 \times 35.00}$

$C_{A} =\frac{3.4229}{105.00}$

$C_{A} = 0.032599 \ M$

$C_{A} \approx 0.0326 \ M$

Hence, the concentration of the unknown H₃PO₄ is 0.0326 M

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