Water Chemistry Problems A water sample was analyzed and was found to have the f

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Water Chemistry Problems

A water sample was analyzed and was found to have the following constituents (same analysis as for part 1 of the homework): Ca+2, mg/L 123

HCO3 - , mg/L 82

Mg+2, mg/L 12

SO4 -2, mg/L 122

Na+ , mg/L 22.1

Cl- , mg/L 112

K+ , mg/L 3.8

CO3 -2, mg/L 6.0

Fe+2, mg/L 5.8

Mn+2, mg/L 1.5 Temperature 25°C

1. Calculate the alkalinity (exactly).

2. Calculate the total, carbonate, and non-carbonate hardness of the water (include contributions made by iron and manganese).

3. How many mL of 0.02N H2SO4 would be required to neutralize the bicarbonate alkalinity in a 50 mL sample?

4. Draw a bar chart for the water (see pages 287-288 for an example).

5. Based on the solubility product for calcium carbonate, how much calcium (mg/L as CaCO3) should be soluble in this water? Is the water under-saturated or over-saturated with respect to calcium?

6. Based on the solubility product for magnesium hydroxide, how much magnesium (mg/L as CaCO3) should be soluble in this water? Is the water under-saturated or over-saturated with respect to magnesium?

Explanation / Answer

To calculate alkalinity you need to apply the next formula:

Alkalinity = [HCO3-] + 2x[CO3(-2)] + [OH-] - [H+] ; formula refers to the concentration of every of these ions

I see you have HCO3 with 82 mg/L

CO3-2 you have 6 mg/L

Alkalinity = 82 + 6 = 88 mg/L

This is normally expressed as Mg/l of CaCO3

to transform HCO3 mg/L to mg/L CaCO3 just divide the value by 1.22 to get a value of 67.21 mg/L as CaCO3

for CO3 you need to divide 6 mg/L by 0.6 to get a value of 10 mg/L as CaCO3

Alkalinity is = 10 + 67.21 = 77.21 mg / L as CaCO3

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