100 mL of a raw water sample required 25 mL of N/50 H2SO4 for neutralisation to phenolphthalein end point. After this methyl orange indicator was added to it and the end point was 28.0 mL. Calculate the alkalinity of water as CaCO3 equvivalents in parts per million.a.Alkalinity due to HCO32– = 60ppm Alkalinity due to CO32– = 220 ppmb.Alkalinity due to OH– = 220 ppm Alkalinity due to CO32– = 60 ppmc.Alkalinity due to HCO32– = 220 ppm Alkalinity due to CO32– = 60 ppmd.Alkalinity due to OH– =60 ppm Alkalinity due to CO32– =220 ppm
Question
100 mL of a raw water sample required 25 mL of N/50 H2SO4 for neutralisation to phenolphthalein end point. After this methyl orange indicator was added to it and the end point was 28.0 mL. Calculate the alkalinity of water as CaCO3 equvivalents in parts per million.a.Alkalinity due to HCO32– = 60ppm Alkalinity due to CO32– = 220 ppmb.Alkalinity due to OH– = 220 ppm Alkalinity due to CO32– = 60 ppmc.Alkalinity due to HCO32– = 220 ppm Alkalinity due to CO32– = 60 ppmd.Alkalinity due to OH– =60 ppm Alkalinity due to CO32– =220 ppm
Solution
The alkalinity of water is determined by the amount of acid needed to neutralize it. In this case, we are given that 100 mL of raw water sample required 25 mL of N/50 H2SO4 for neutralization to the phenolphthalein end point. This indicates the presence of bicarbonates (HCO3-) and carbonates (CO3 2-) in the water.
-
First, we calculate the total alkalinity. The total alkalinity is given by the volume of acid used to the methyl orange end point. This is given as 28 mL of N/50 H2SO4.
Total alkalinity = (28 mL * N/50 * 1000 mg/L) / 100 mL = 560 mg/L as CaCO3
-
Next, we calculate the phenolphthalein alkalinity. This is given by the volume of acid used to the phenolphthalein end point. This is given as 25 mL of N/50 H2SO4.
Phenolphthalein alkalinity = (25 mL * N/50 * 1000 mg/L) / 100 mL = 500 mg/L as CaCO3
-
The difference between the total alkalinity and the phenolphthalein alkalinity gives the alkalinity due to bicarbonates (HCO3-).
Alkalinity due to HCO3- = Total alkalinity - Phenolphthalein alkalinity = 560 mg/L - 500 mg/L = 60 mg/L as CaCO3
-
The phenolphthalein alkalinity is due to the presence of carbonates (CO3 2-).
Alkalinity due to CO3 2- = Phenolphthalein alkalinity = 500 mg/L as CaCO3
Therefore, the alkalinity of the water as CaCO3 equivalents in parts per million is 60 ppm due to HCO3- and 500 ppm due to CO3 2-.
So, the correct answer is: Alkalinity due to HCO3- = 60 ppm, Alkalinity due to CO3 2- = 500 ppm.
Similar Questions
A water sample was alkaline to both phenolphthalein and methyl orange. 50 mL of the watersample required 18 mL of 0.02 N H2SO4 for phenolphthalein end point and another 6 mLfor complete neutralisation. Describe the type and amount of alkalinity present.a.Alkalinity due to OH– = 24 ppm, Alkalinity due to CO32– = 24 ppmb.Alkalinity due to OH– = 240 ppm, Alkalinity due to CO32– = 240 ppmc.Alkalinity due to OH– = 340 ppm, Alkalinity due to CO32– = 140 ppmd.Alkalinity due to OH– = 00 ppm, Alkalinity due to CO32– = 480 ppmClear my choice
100 ml of an alkaline water in presence of phenolphthalein, 50 ml of the N/100 HCl. The resulting mixture in presence of methyl orange indicator consumes 35 ml of HCl. Predict the alkalinities and calculate their amounts in ppm as CaCO3 equivalent.
A sample of water was alkaline both to phenolphthalein and methyl orange. 100 ml of this water sample required 20 ml of N/50 H2SO4 for phenolphthalein end point and another 15 ml for complete neutralization. Calculate the type and amount of alkalinity in ppm.
A water sample is alkaline to both phenolphthalein as well as methyl orange. 200 mL of a this water sample on titration with N/50 HCl required 9.4 mL of the acid to phenolphthalein end point. When a few drops of methyl orange are added to the same solution and titration is further continued, the yellow color of the solution just turned red after addition of 30.4 mL of the acid solution. Calculate the alkalinity present in the water sample.a.P = 152 ppm., M = 47ppm and HCO3– alkalinity = 58 ppm, CO32– alkalinity = 94 ppmb.P = 1520 ppm., M = 470ppm and HCO3– alkalinity = 940 ppm, CO32– alkalinity = 580 ppmc.P = 47 ppm., M = 152 ppm and HCO3– alkalinity = 58 ppm, CO32– alkalinity = 94 ppmd.P = 470 ppm., M = 1520 ppm and HCO3– alkalinity = 580 ppm, CO32– alkalinity = 940 ppm
If a solution of pure water and sodium hydroxide is titrated with H2SO4 for the alkalinity measurement. Then the correct option would be...a.The Phenolphthalein alkalinity will be the total alkalinityb.The Phenolphthalein alkalinity will be the bicarbonate alkalinityc.The Methyl orange alkalinity will be the total alkalinityd.The Phenolphthalein alkalinity will not be the total alkalinity
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.