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Andrews [41]
2 years ago
9

Calculate the volume in mL of 0.279 M Ca(OH)2 needed to neutralize 24.5 mL of 0.390 M H3PO4 in a titration.

Chemistry
1 answer:
Vsevolod [243]2 years ago
5 0

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

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8 0
2 years ago
“Dry ice” is the solid form of carbon dioxide. Determine the number of mass of CO2 gas in grams that are present in 61.8 L of CO
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Answer:

Mass  = 121 g

Explanation:

Given data:

Mass in gram of CO₂ = ?

Volume = 61.8 L

Pressure = standard = 1 atm

Temperature = 273.15 K

Solution:

Formula:

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

1 atm × 61.8 L = n ×0.0821 atm.L/ mol.K   × 273.15 k

61.8 L.atm = 22.42 atm.L/ mol × n

n = 61.8 L.atm /22.42 atm.L/ mol

n = 2.76 mol

Mass in gram:

Mass =  number of moles × molar mass

Mass = 2.76 mol × 44 g/mol

Mass  = 121 g

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