Question

A groundwater contains the following cations (expressed as the cation):
Na+ = 20 mg/L
Ca++ =50 mg/L
Mg++ = 20 mg/L
The total hardness (expressed as mg/L as CaCO3) is most nearly:

Answer 1

Answer:

208 mg/L

Explanation:

Only Mg++ and Ca++ causes hardness not Na+

Given

Mg++ = 20 mg/L

Ca++ = 50 mg/L

$Hardness\ of\ water = C^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ C^{2+}}\ +\ mg^{2+}\times \frac{equivalent\ weight\ of\ CaCO3}{equivalent\ mass\ of\ mg^{2+}}$

= $50\times\frac{50}{20}\ +\ 20\times\frac{50}{12}$

= 208.33 mg/L $\approx$ 208mg/L

Answer 2

Answer:

50mg/L

Explanation:

CaCO3 -> Ca++ + CO2