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
1 answer:
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 = = 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>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 =
Therefore, the new concentration of the H₃PO₄ solution is 0.0432 mol/L
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Explanation:
As the element has 5 valence electrons in the sixth energy level, which mens it belongs to 6th period as the valence electron has entered the 6th energy level.
Now as the electrons are filled according to afbau's rule , the electronic configuration of the element will be :
Now as the atomic number of element is equal to the number of electrons in a neutral atom , the atomic number is 83 and the element is Bismuth (Bi). As the last electron enters p -subshell , it is a p block element. It lies in group 15 which is nitrogen family.