Question

If 42.8 mL of 0.200 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?

Answer 1

Here, HCl is a strong monobasic acid and Ca(OH)2 is a strong mono acidic base. So, there is a complete ionization of both in water.
From, balanced equation:
2HCl + Ca(OH)2 --> CaCl2 + 2H2O

2 moles of HCl molecules neutralize only one mole of Ca(OH)2.

This indicates that the concentration of Ca(OH)2 would be half of the concentration of HCl. Now, first find out the concentration of HCl as:
The moles of solute HCl in 42.8 mL solution:

Molarity = moles / L (volume)

0.200M = moles/(0.0428 L) [as 1L = 1000mL]
So, moles = 0.00848

Now, divide this by 2 to which will provide the moles of Ca(OH)2:
0.00848/2 = 0.00422 moles
This the value of Ca(OH)2 in moles as solute in the solutio of Ca(OH)2

Finally, from relation:
moles of solute = mass of solute in gram / molar mass of solute
[molar mass of Ca(OH)2 = 74 g/mole]
So, mass of solute Ca(OH)2 = moles of solute Ca(OH)2 x molar mass of Ca(OH)2
= 0.00422 x 74
= 0.3122g of Ca(OH)2