  Question

# 20 liter of tap water that hasCa(HCO3)2 was treated with 90 ml of M/50 HCl. The resulting solution requires an extra 6 ml of 0.1 M NaOH to make the solution neutral. What is the total hardness as parts of CaCO3 in ppm? (Take the molar mass of Ca(HCO3)2 = 162 g/mol)

Solution

## The acid neutralises the Ca(HCO3)2, extra acid remains which makes the solution acidic. The base added neutralises this. Since moles react, let's first find out the moles of the acid , base and Ca(HCO3)2 involved. We will then figure out how many moles of HCl were used by Ca(HCO3)2 and then find out the hardness of the water in ppm using the molar mass of CaCO3. Also, ppm is same as mg/L. So we need to find out the mass of CaCO3 in 1 L of the solution. This last step is the most important trick here. Number of moles = molarity × volume. Since the volume is in mL, we get millimoles (mmol). 90 ml of M/50 HCl=1.8 mmol of HCl 6 ml of 0.1 M NaOH = 0.6 mmol of NaOH HCl used to neutralise Ca(HCO3)2 = 1.8 - 0.6 = 1.2 mmol Since 1 mole of Ca(HCO3)2 requires 2 moles of the acid to get neutralised,  Ca(HCO3)2 present =1.2/2 = 0.6 mmol. This is present in 20 L of water.  1 L of water would have 0.6/20 = 0.03 mmol of Ca(HCO3)2 So, mass of CaCO3 = 0.03×100 = 3 mg/L ≡ 3 ppm.  Suggest corrections   