7 a sample of water was alkaline to both phenolphthalein and...

7. A sample of water was alkaline to both Phenolphthalein and methyl orange. 25 mL of this water sample required 15 mL of 0.02 N HCl for the phenolphthalein end point and another 10 mL for complete neutralization. Calculate the type of alkalinity.

Transcript

00:01 Hello, in this question we need to find out the alkali -nity, that is alkyne it is there, alkylaity of water.

00:16 So here, as we know that, that is 100 ml of water up to phenotheline endpoint, it is, it is equal to 7 .5 ml of n by 50 of acid.

00:39 H2so4.

00:40 So therefore here 100 multiplied by this is the strength is equal to 7 .5 multiplied by n by 50.

00:53 So therefore here this n p it is equal to this will become 7 .5 divided by 100 multiplied by n divided by 50.

01:04 So strength of alkalinity up to fiddle.

01:09 An ophthalene and point in terms of c .a.

01:13 Co3.

01:15 So here we have strength.

01:19 It is equal to np multiply by equivalent weight of c .a.

01:27 Co3.

01:28 So here, this np, it is equal to 7 .5 divided by 100.

01:34 Multiply by n value here, it is 1 divided by 50, multiply by 50, multiply by 50 gram per liter.

01:41 So, here it is in the gram so therefore we can write it as 7 .5 divided by 100 multiplied by 50 multiplied by 50 multiply by here.

01:55 This gram can be converted into milligram, thousand milligrams per liter because one gram it is equal to 10 days to the power 3 milligram.

02:06 So 50 to 50 got cancelled out and 2 0 and 200 and therefore we will have 75.

02:12 Milligram per liter or we can write it as 75 parts per million so as 100m .m.

02:23 Of water 100m.

02:27 Of water up to methyl orange point it is equal to 7 .5 plus 10m...

Question

7. A sample of water was alkaline to both Phenolphthalein and methyl orange. 25 mL of this water sample required 15 mL of 0.02 N HCl for the phenolphthalein end point and another 10 mL for complete neutralization. Calculate the type of alkalinity.

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