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3 years ago

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A titration reaction requires 38.20 mL phosphoric acid solution to react with 71.00 mL of 0.348 mol/L calcium hydroxide to reach

the endpoint. Based on this information, determine the following:
a. Balance equation: Answer____
Ca(OH)2 (aq) + Answer_____
H3PO4 (aq) ------> Answer____
(aq) + Answer___
H2O

b. Number of moles ( to 4dp) of Ca(OH)2 : Answer_____

c. Number of moles ( to 4dp) of H3PO4 : Answer____

d. Concentration of H3PO4 (to 3dp): Answer____-

ii. If a 10ml aliquot of the H3PO4 solution was diluted to 100ml. What would be the new concentration of the solution (to 4dp) ?

1 answer:

NARA [144]3 years ago

3 0

1a. The balanced equation for the reaction is:

<h3>3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O </h3>

1b. The number of mole of Ca(OH)₂ is 0.0247 mole

1c. The number of mole of H₃PO₄ is 0.0165 mole.

1d. The concentration of H₃PO₄ is 0.432 mol/L

2. The new concentration of the H₃PO₄ solution is 0.0432 mol/L

<h3>1a. The balanced equation for the reaction</h3>

<u>3</u>Ca(OH)₂ + <u>2</u>H₃PO₄ —> Ca₃(PO₄)₂ + <u>6</u>H₂O

<h3>1b. Determination of the mole of Ca(OH)₂</h3>

Volume of Ca(OH)₂ = 71 mL = 71 / 1000 = 0.071 L

Concentration of Ca(OH)₂ = 0.348 mol/L

<h3>Mole of Ca(OH)₂ =? </h3>

Mole = Concentration × Volume

Mole = 0.348 × 0.071

<h3>Mole of Ca(OH)₂ = 0.0247 mole </h3>

<h3>1c. Determination of the mole of H₃PO₄. </h3>

3Ca(OH)₂ + 2H₃PO₄ —> Ca₃(PO₄)₂ + 6H₂O

From the balanced equation above,

3 moles of Ca(OH)₂ reacted with 2 moles of H₃PO₄.

Therefore,

0.0247 moles of Ca(OH)₂ will react with = $\frac{0.0247 * 2}{3}$ = 0.0165 mole of H₃PO₄.

Thus, the number of mole of H₃PO₄ is 0.0165 mole

<h3>1d. Determination of the concentration of H₃PO₄</h3>

Volume of H₃PO₄ = 38.20 mL = 38.20/ 1000 = 0.0382 L

Mole of H₃PO₄ = 0.0165 mole

<h3>Concentration of H₃PO₄ =?</h3>

$Concentration = \frac{mole}{volume} \\\\Concentration = \frac{0.0165}{0.0382}$

<h3>Concentration of H₃PO₄ = 0.432 mol/L</h3>

<h3>2. Determination of the new concentration of the H₃PO₄ solution.</h3>

Initial Volume (V₁) = 10 mL

Initial concentration (C₁) = 0.432 mol/L

New volume (V₂) = 100 mL

<h3>New concentration (C₂) =?</h3>

The new concentration of the H₃PO₄ solution can be obtained as follow:

<h3>C₁V₁ = C₂V₂</h3>

0.432 × 10 = C₂ × 100

4.32 = C₂ × 100

Divide both side by 100

C₂ = $\frac{4.32}{100}\\$

<h3>C₂ = 0.0432 mol/L</h3>

Therefore, the new concentration of the H₃PO₄ solution is 0.0432 mol/L

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