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Hi, I need help understanding \"Hard Water\" with measuring with calcium and mag

ID: 987114 • Letter: H

Question

Hi, I need help understanding  "Hard Water" with measuring with calcium and magnesium. Can i get a step by step explanation how it works and if there is an equation that can be used to calculate it to help understand the relationship of ion and hard water. Thanks for the help! Hi, I need help understanding  "Hard Water" with measuring with calcium and magnesium. Can i get a step by step explanation how it works and if there is an equation that can be used to calculate it to help understand the relationship of ion and hard water. Thanks for the help!

Explanation / Answer

Hard water is the concept used to measure concnetration of Ca+2 and Mg+2 ions

these two will form precipitate when heating (which is not common, since solubilty of plenty of salts increase when T increases)

Therefore;

the "hardness" of water is a measurement of how much Ca+2 ions are present

This is the typicall reaction in equilibrium, which is changed as T is modified

CaCO3 (s) + CO2 (aq) + H2O (l) Ca2+ (aq) + 2HCO3 (aq)

Hardwater is measured typically as:

ppm (mg of Ca+2 ions vs. kg/Liter of solution)

mmol/L (milimol of Ca+2 ions per liter of solution)

Therefore

if we have

a hardwater with 40 ppm

this means, there is 40 parts per million of Ca+2 (million since mg = 10^-3 and kg = 10^3; therefore 10^6 overall)

40 mg of Ca+2 per kg of solution

in this case the solution is mainly water so

40 mg of Ca+2 per kg of water

typically, 1 kg of water = 1 L of water so

40 mg of Ca+2 ions per liter of water

If it is expressed in mmol

has a use in reaction and stoichiometric values

1 mmol of Ca+2 will react with 2 mmol of HCO3- to form 1 mmol of CaCO3 (easy to understand in milimol rather than masses)

therefore

if we have a 2.5 mmol concentration of water then

2.5 mmol of Ca+2 is present in 1 liter of solution

2.5*10^-3 mol of Ca+2

mass = mol*MW

MW of Ca+2 = 40 g/mol so

mass = (2.5*10^-3)(40) = 0.1 g of Ca+2

mass in mg = 0.1*1000 = 100 mg of Ca+2 per liter of soluion

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