Question

100 ml of a sample of hard water requires 20 ml of 0.03 N H\textsubscript{2}SO\textsubscript{4} for complete reaction. The hardness of water (density = 1 gm/ml) is:

Answer 1

300 mg/L

Answer 2

Solution:

1. Calculation of equivalents consumed by acid:

   Volume of H\textsubscript{2}SO\textsubscript{4} = 20 mL = 0.02 L

   Normality = 0.03 N

   Equivalents in 100 mL water = 0.02 × 0.03 = 0.0006 eq

2. Extrapolate to 1 L of water:

   Since 100 mL has 0.0006 eq, 1 L has

   0. 0006 × 10 = 0.006 eq

3. Convert equivalents to mg CaCO\textsubscript{3} (using CaCO\textsubscript{3} equivalent weight = 50 mg/meq):

   Hardness = 0.006 eq/L × 1000 mg/eq (because 1 eq of CaCO\textsubscript{3} = 50 g = 50,000 mg divided by 1000 if 50 mg per meq)

   Alternatively, in meq: 0.006 eq = 6 meq.

   Hardness (mg CaCO\textsubscript{3}) = 6 meq/L × 50 mg/meq = 300 mg/L

Explanation:

Calculate equivalents of acid used in 100 mL sample, scale to 1 L (multiply by 10) to get 6 meq/L, then multiply by 50 (mg/ meq conversion factor) to determine hardness as 300 mg/L CaCO\textsubscript{3}.