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if 27.5 mL of 0.125 M HCl solution is needed to neutralize a solution of CA(OH)2 how many grams of CA(OH)2 must be in the solution?

User Mahalie
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1 Answer

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Step-by-step explanation:

The reaction between HCl and Ca(OH)₂ is:

2 HCl + Ca(OH)₂ ----> CaCl₂ + 2 H₂O

First we have to find the number of moles of HCl that were needed to neutralize the Ca(OH)₂ solution. We used 27.5 mL of a 0.125 M solution.

volume of solution in L = 27.5 mL * 1 L/1000 mL

volume of solution in L = 0.0275 L

moles of HCl = 0.125 moles/L * 0.0275 L

moles of HCl = 0.003438 moles

According to the coefficients of the reaction, 2 moles of HCl will neutralize 1 mol of Ca(OH)₂. The molar ratio between them is 2 to 1. We can use this relationship to find the number of moles of Ca(OH)₂ that will be neutralized by 0.003438 moles of HCl.

2 moles of HCl = 1 mol of Ca(OH)₂

moles of Ca(OH)₂ = 0.003438 moles of HCl * 1 mol of Ca(OH)₂/(2 moles of HCl)

moles of Ca(OH)₂ = 0.001719 moles

Finally we can convert these moles into grams using the molar mass of Ca(OH)₂.

molar mass of Ca(OH)₂ = 74.09 g/mol

mass of Ca(OH)₂ = 0.001719 moles * 74.09 g/mol

mass of Ca(OH)₂ = 0.127 g

Answer: the mass of Ca(OH)₂ that must be in the solution is 0.127 g.

User Venkatesh G
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