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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

Question

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

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Solution

The alkalinity of a water sample can be determined by titration with a standard acid to designated pH levels. In this case, the water sample is alkaline to both phenolphthalein and methyl orange, which means it contains hydroxides (OH-) and/or carbonates (CO32-).

The phenolphthalein alkalinity (P) is determined by titration to a pH of 8.3, which will neutralize both the hydroxides and half of the carbonates. The total alkalinity (T) is determined by titration to a pH of 4.5, which will neutralize the hydroxides, carbonates, and bicarbonates.

Given that 50 mL of the water sample required 18 mL of 0.02 N H2SO4 for phenolphthalein end point, we can calculate the phenolphthalein alkalinity (P) as follows:

P = (Volume of H2SO4 for P end point) x (Normality of H2SO4) x (Equivalent weight of CaCO3) x (1000 / Sample volume) P = (18 mL) x (0.02 N) x (50 g/mol) x (1000 / 50 mL) P = 360 mg/L or 360 ppm

Given that the water sample required another 6 mL of 0.02 N H2SO4 for complete neutralisation, we can calculate the total alkalinity (T) as follows:

T = (Total volume of H2SO4) x (Normality of H2SO4) x (Equivalent weight of CaCO3) x (1000 / Sample volume) T = (18 mL + 6 mL) x (0.02 N) x (50 g/mol) x (1000 / 50 mL) T = 480 mg/L or 480 ppm

The alkalinity due to hydroxides (OH-) is equal to the phenolphthalein alkalinity (P), and the alkalinity due to carbonates (CO32-) is equal to the total alkalinity (T) minus the phenolphthalein alkalinity (P). Therefore:

Alkalinity due to OH- = P = 360 ppm Alkalinity due to CO32- = T - P = 480 ppm - 360 ppm = 120 ppm

So, none of the given options is correct. The correct answer should be:

Alkalinity due to OH- = 360 ppm, Alkalinity due to CO32- = 120 ppm

This problem has been solved

Similar Questions

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

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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

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.

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