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

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

If the carbonate ion is only responsible for the alkalinity, thena.The phenolphthalein alkalinity would be greater than half of the methyl orange alkalinityb.The phenolphthalein alkalinity would be equal to half of the methyl orange alkalinityc.The phenolphthalein alkalinity would be equal to the methyl orange alkalinityd.The phenolphthalein alkalinity would be lower than half of the methyl orange alkalinityClear my choice

If the phenolphthalein alkalinity is equal to the total alkalinity then what would be the reason behind alkalinity?a.Bicarbonate ions onlyb.Carbonate ions onlyc.Both, hydroxide and carbonate ionsd.Hydroxide ions onlyClear my choice

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

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