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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

Question

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

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Solution

The alkalinity of a water sample can be determined by titration with a standard solution of a strong acid, usually HCl. In this case, the water sample is alkaline to both phenolphthalein and methyl orange, which means it contains both bicarbonate (HCO3-) and carbonate (CO32-) ions.

The volume of HCl required to reach the phenolphthalein endpoint (P) is 9.4 mL. This represents the total alkalinity due to the presence of both HCO3- and CO32- ions.

The volume of HCl required to reach the methyl orange endpoint (M) is 30.4 mL. This represents the total alkalinity due to the presence of HCO3-, CO32-, and hydroxide (OH-) ions.

The difference between M and P (M - P) represents the alkalinity due to the presence of OH- ions.

The alkalinity due to HCO3- is equal to P, the alkalinity due to CO32- is equal to (M - P)/2, and the alkalinity due to OH- is equal to (M - P)/2.

To convert these volumes to ppm, we use the formula:

ppm = (mL HCl * N HCl * equivalent weight of the ion * 1000) / mL sample

The equivalent weight of HCO3- is 61, of CO32- is 30, and of OH- is 17.

So, the alkalinity due to HCO3- is (9.4 mL * 50 * 61 * 1000) / 200 mL = 1415 ppm.

The alkalinity due to CO32- is ((30.4 mL - 9.4 mL) / 2 * 50 * 30 * 1000) / 200 mL = 1580 ppm.

The alkalinity due to OH- is ((30.4 mL - 9.4 mL) / 2 * 50 * 17 * 1000) / 200 mL = 899 ppm.

So, the total alkalinity is 1415 ppm + 1580 ppm + 899 ppm = 3894 ppm.

Therefore, the answer is not listed among the options provided.

This problem has been solved

Similar Questions

A sample of water was alkaline to both phenolphthalein and methyl orange.50 ml of this water sample required 15 ml of N/50 HCl for phenolphthalein end point and another 10 ml for complete neutralisation in the same solution. Calculate the type and extent of alkalinity in ppm.

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 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.

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

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

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