Question

Identify the pair of chlorides responsible for permanent hardness of water.

• A. NaCl, KCl
• B. CaCl\(_2\), KCl
• C. AlCl\(_3\), MgCl\(_2\)
• D. MgCl\(_2\), CaCl\(_2\)

Hint:

A simple rule to remember for water hardness:
Hardness Ions: Ca\(^{2+}\) and Mg\(^{2+}\).
Anions for Temporary Hardness: Bicarbonate (HCO\(_3^-\)).
Anions for Permanent Hardness: Chloride (Cl\(^-\)) and Sulfate (SO\(_4^{2-}\)).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are required to determine the degree of hardness of the given water sample in ppm.

Step 2: Key Formula or Approach:
Convert MgCl₂ into its equivalent amount of CaCO₃ and then calculate ppm based on the water sample mass.

Step 3: Detailed Explanation:
Given 19 mg of MgCl₂ in 2 kg of water, the equivalent hardness as CaCO₃ is \(19\times100/95=20\) mg. Dividing 20 mg by 2 kg gives 10 mg/kg. Since 1 mg/kg corresponds to 1 ppm, the hardness of the sample is 10 ppm.

Step 4: Final Answer:
The hardness of the given sample is 10 ppm.