2HCO3 - + Ca2+ CaCO3 + CO2 + H2O Bicarbonate (HCO3-) combines with calcium ions in the water to make calcium carbonate (CaCO3, limestone). This process can occur both within organisms such as corals or as a simple chemical reaction in the water itself.
The volume of the 0.279 M Ca(OH)₂ solution required to neutralize 24.5 mL of 0.390 M H₃PO₄ is 51.4 mL
<h3>Balanced equation </h3>
2H₃PO₄ + 3Ca(OH)₂ —> Ca₃(PO₄)₂ + 6H₂O
From the balanced equation above,
- The mole ratio of the acid, H₃PO₄ (nA) = 2
- The mole ratio of the base, Ca(OH)₂ (nB) = 3
<h3>How to determine the volume of Ca(OH)₂ </h3>
- Molarity of acid, H₃PO₄ (Ma) = 0.390 M
- Volume of acid, H₃PO₄ (Va) = 24.5 mL
- Molarity of base, Ca(OH)₂ (Mb) = 0.279 M
- Volume of base, Ca(OH)₂ (Vb) =?
MaVa / MbVb = nA / nB
(0.39 × 24.5) / (0.279 × Vb) = 2/3
9.555 / (0.279 × Vb) = 2/3
Cross multiply
2 × 0.279 × Vb = 9.555 × 3
0.558 × Vb = 28.665
Divide both side by 0.558
Vb = 28.665 / 0.558
Vb = 51.4 mL
Thus, the volume of the Ca(OH)₂ solution needed is 51.4 mL
Learn more about titration:
brainly.com/question/14356286
Answer:
313, 6grams of H3PO4
Explanation:
We calculate the weight of 1 mol of H3PO4:
Weight 1 mol H3PO4= (Weight H)x3+ (Weight P)+(Weight 0)x4 =1gx3+31g+16gx4
Weight 1 mol H3PO4=98 g /mol
1 mol-----98 grams H3PO4
3,2mol----x= (3,2molx 98 grams H3PO4)/ 1mol=313,6 grams H3PO4
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