A water sample was analyzed and was found to have the following constituents (sa

Question

A water sample was analyzed and was found to have the following constituents (same analysis as for part 1 of the homework): Ca2, mg/L 123 Mg*, mg/L 12 Na, mg/L 22.1 K', mg/L3.8 Fe 2, mg/L 5.8 Mn*2, mg/L 1.5 HCO, mg/L 82 SO,2, mg/L 122 CI, mg/L 112 co,2, mg/L 6.0 Temperature 25oC 1.Calculate the alkalinity (exactly) 2. Calculate the total, carbonate, and non-carbonate hardness of the water (include contributions made by iron and manganese). How many mL of 0.02N H SO4 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 CaCO,) 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 CaCO,) should be soluble in this water? Is the water under-saturated or over-saturated witlh 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/ml 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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