Question

calculate the temporary and permanent hardness for the water containing MgSo4-25 mg/lit Mg(HCO3)2- 32 mg/litre cacl2 55mg/litre​

Answer 1

Answer:

Temporary hardness = Mg(HCO3)2 = 11.50mgs/Lit. Permanent hardness = MgCl2 + MgSO4 = 12.63 + 24.66 = 37.29mgs/Lit.

The total hardness is determined by the sum of Ca and Mg: TH = (2×1.40 + 2×0.23) meq/L = 3.26 meq/L –

Answer 2

Answer:

The temporary hardness due to $Mg(HCO_{3} )_{2}$ is $21.91mg/L$.

The permanent hardness due to $MgSO_{4}$ and $CaCl_{2}$ is $70.37mg/L$.

Explanation:

Given,

The weight of $MgSO_{4}$ = $25mg/L$

The weight of $Mg(HCO_{3})_{2}$ = $32mg/L$

The weight of $CaCl_{2}$ = $55mg/L$

The temporary hardness =?

The permanent hardness =?

Firstly, we have to find out the temporary hardness.

As we know,

• The temporary hardness is caused mainly by the presence of dissolved bicarbonates of Ca, Mg, and other heavy metals.

So, in this case, the temporary hardness is caused by $Mg(HCO_{3})_{2}$.

Now, we have to convert its weight into the equivalents of $CaCO_{3}$.

For this, we have to multiply its weight by the formula given below:

• $CaCO_{3}$ equivalent = $\frac{Molar mass of CaCO_{3} }{Molar mass of hardness causing substance}$
• The molar mass of $Mg(HCO_{3})_{2}$ = $146g/mol$
• The molar mass of $CaCO_{3}$ = $100g/mol$

Therefore,

• $CaCO_{3}$ equivalent for $Mg(HCO_{3})_{2}$ = $\frac{100}{146}$
• $CaCO_{3}$ equivalent for $32mg/L$ $Mg(HCO_{3})_{2}$ = $32\times \frac{100}{146}$ = $21.91mg/L$

Therefore,

• The temporary hardness of water due to $Mg(HCO_{3} )_{2}$ = $21.91mg/L$.

Now, we have to calculate the permanent hardness.

As we know,

• The permanent hardness is caused by the presence of chlorides and sulfates of Ca, Mg, and other heavy metals.

So, in this case, the temporary hardness is caused by $MgSO_{4}$ and $CaCl_{2}$.

Now, we have to convert their weights into equivalents of $CaCO_{3}$.

For this, we have to multiply its weight by the formula given below:

• $CaCO_{3}$ equivalent = $\frac{Molar mass of CaCO_{3} }{Molar mass of hardness causing substance}$

For $MgSO_{4}$,

• The molar mass of $MgSO_{4}$ = $120g/mol$
• $CaCO_{3}$ equivalent for $MgSO_{4}$ = $\frac{100}{120}$
• $CaCO_{3}$ equivalent for $25 mg/L$ $MgSO_{4}$ = $25\times \frac{100}{120}$ = $20.83mg/L$

For $CaCl_{2}$,

• The molar mass of $CaCl_{2}$ = $111g/mol$
• $CaCO_{3}$ equivalent for $CaCl_{2}$ = $\frac{100}{111}$
• $CaCO_{3}$ equivalent for $55mg/L$ $CaCl_{2}$ = $55\times \frac{100}{111}$ = $49.54mg/L$

Therefore,

• The permanent hardness due to $MgSO_{4}$ and $CaCl_{2}$ = $20.83 + 49.54$ = $70.37mg/L$.