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:
See below.
Explanation:
pH = - log (H+)
So the pH of 1 x 10^-7 M solution is 7.
I'm sorry but I'm not sure about what the other units mean, so Im not sure of the answer to those.
If you convert the other units to the form 1 x 10^-n then the pH will be n.
The answer is 3. As 5 * 3 = 15.
Is there any other equations I may be able to help you with? :)
Answer:
In general an acid reacts with a carbonate or hydrogen-carbonate to produce a salt, carbon dioxide gas and water.
Answer:Mass of Potassium chloride =1.762g
Explanation:
Mass of empty beaker = 23.100 g
Mass of beaker with Potassium chloride = 24.862g
Mass of Potassium chloride = Final weight - initial weight = Mass of beaker with Potassium chloride - Mass of empty beaker = 24.862-23.100 = 1.762g
