Calculate Temporary and Permanent Hardness of Water (NEET Chemistry)

Calculate the temporary and permanent hardness of water sample having the following the following constituents per litre:

Ca(HCO₃)₂=162mg ,MgCl₂=95=mg

NaCl=585mg , Mg(HCO3)₂=73mg

CaSO₄=136mg

A.200ppm, 150ppm

B.100ppm,150ppm $$\mathrm{}$$

C.150ppm 200ppm

D.150ppm, 150ppm

Hint:

-Permanent hardness cannot be removed by simple boiling, whereas temporary hardness can be removed by simple boiling.
-To calculate the temporary and permanent hardness of a sample we need to calculate the total number of moles of calcium and magnesium, which is present in the ions of salt.

Complete step by step answer:
The most commonly observed difference between soft and hard water is that hard water reacts with soap in order to form scum of solid soap, which is nothing but the white lather type substance which forms as a precipitate, due to presence of calcium and magnesium ions in the hard water, and soft water does not form scum of soap. The formation of soap scum on your skin is the main reason behind the fact that it is easier to rinse from your body while using soap, when you shower with hard water.

Hardness in water is produced due to the presence of magnesium ions, calcium ions or both. With increase in concentrations of magnesium ions, calcium ions or both, hardness of water also increases. The combined concentration of magnesium and calcium observed in a Solution is frequently referred to as total hardness.

When hard water is exposed to a higher temperature or in other words when it is boiled in the presence of bicarbonate (HCO₃) or carbonate (CO₃)²⁻the magnesium and/or calcium ions react with the bicarbonate and carbonate ions in order to form solid magnesium carbonate, calcium carbonate, or both.

Temporary hardness is generally more complex, because concentration of it is a function of the concentration of carbonates in relation to their reaction with calcium in magnesium.
Permanent hardness is a type of hardness which simply cannot be removed by boiling of water.

According to the question,

$$\mathrm{mole\; of\; C}a{\left(HC{O}_3\right)}_2={\displaystyle \frac{162\times {10}^{-3}mg}{162g/mole}}=1\times {10}^{-3}moles$$

$$\mathrm{Mole\; of\; C}a\left(S{O}_4\right)={\displaystyle \frac{136\times {10}^{-3}g}{136g/mole}}=1\times {10}^{3}mole$$

$$\mathrm{We\; know\; that\; total\; mole\; of \; C}a=2\times {10}^{-3}mole$$

Now we will calculate mole of

$$MgC{l}_2={\displaystyle \frac{95\times {10}^{-3}}{95}}=1\times {10}^{-3}mole$$ $$\mathrm{Mole\; M}g={\displaystyle \frac{73\times {10}^{-3}}{146}}=5\times {10}^{-4}mole$$

So the correct option is C.

Note: The extent of permanent and temporary hardness can be calculated by simply calculating the number of moles of calcium and magnesium present in the sample Solution which contains calcium bicarbonate, calcium sulphate as well as magnesium bicarbonate and magnesium chloride along with sodium chloride whose values where provided in the question.