The solubility of a slaked lime, Ca(OH)2, in water is 0.185g/100.0 mL. What volume of 2.95×10-3 M HCl is needed to neutralize 13.0 mL of a saturated Ca(OH)2 solution? Answer is in mL.

Answer 1

Since the solubility is 0.185 g / 100 mL, the amount in one mL is 0.185/ 100 = 0.00185 g, and the mass in 13 mL is 13 x 0.00185 = 0.02405 g. This calculates to 0.02405 g / 74.092 g/mole = 3.246 x 10^-4 moles of Ca(OH)2.

The balanced equation is Ca(OH)2 + 2HCl ---> CaCl2 + 2H2O. Since we need 2 moles of HCl for every mole of Ca(OH)2, we will need 2 x 3.246 x 10^{-4 =} = 6.492 x 10^-4 moles of HCl.

2.95 x 10^-3 M HCL has 2.95 x 10^-3 /1000 mL, so 1 mL contains 2.95 x 10^-6 moles HCl.. Use the calculation

1 mL / 2.95 x 10^-6 moles x 6.492 x 10^-4 moles HCl = 220 mL HCl solution needed.

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