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A 50.00 mL sample of aqueous Ca(OH)2 requires 34.66 mL of a 0.944M nitric acid for...

A 50.00 mL sample of aqueous Ca(OH)2 requires 34.66 mL of a 0.944M nitric acid for neutralization. Calculate the concentration (molarity) of the original solution of calcium hydroxide.

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Answer #1

V1S1 = V2S2

Where

V1 = volume of Ca(OH)2

S1 = Strength of Ca(OH)2 in normality

V2 = volime of HNO3

S2 = strength of HNO3 in normality

Therefore S1 = V2S2 / V1 = 34.66 x 0.944 / 50.00    [as 0.944 M HNO3 ? 0.944 N HNO3]

Or S1 = 0.654 N

Now 0.654 N Ca(OH)2 ? (0.654/ 2) M Ca(OH)2                     [as equivalent wt of Ca(OH)2 = mol. Wt /2]

                                    = 0.327 M

Therefore concentration of original solution of calcium hydroxide is 0.327 moles/ Lt.

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