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What is the total hardness of sample of water has the following impurities in mg/I. Ca(HCO3)2 = 162 CaCI2 = 22.2 MgCI2 = 95 NaCI = 20
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Given:

$\begin{align*} Ca(HCO_3)_2&=162\\ CaCl_2&=22.2\\ MgCl_2&=95\\ NaCl&=20 \end{align*}$

Solution:

$\text{Equivalents of Calcium Carbonate}=\dfrac{\text{Mass of hardness producing substance}\times50} {\text{Chemical equivalent of hardness producing substance}}$

1.

$Ca(HCO_3)_2=\dfrac{162\times 50}{162.11}=49.966\ mg/L$

2.

$CaCl_2=\dfrac{22.2\times 50}{110.98}=10\ mg/L$

3.

$MgCl_2=\dfrac{95\times 50}{95.21}=49.889\ mg/L$

4.

$NaCl=\dfrac{20 \times 50}{40}=25\ mg/L$

 

$\begin{align*} \text{Total Hardness}&=49.97+10+49.89+25\\ &=134.86\ mg/L \end{align*}$

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