00:03
Here, to calculate the amount of magnesium to positive present in sea water, we need to use ksp.
00:10
First of all, we need to determine the mass of mg, which is 1 .0, 10 raised 2 par 3 liters, multiplied with 1 ,000 centimeter cube per liter, 6 gram per centimeter cube, multiplying with 1 to 7 2 hours per million this gives the mass as 1 .305 10 days to per 3 grams the concentration is this if 99 .9 is to be recovered that means 0 .9991 .999 1 .9 .9 .9 .1 .1 .1 .1 .1 .1 .1.
01:14
1 .305 grams per liter will be equal to 1 .304 grams per liter as per as 99 .9 % the molar concentration can be calculated as divided by 24 .305 gram per mole and for gram per liter.
01:43
The molar concentration turn out to be 0 .0 5365 molar.
01:50
As for calcium hydroxide reaction with magnesium on one ratio 1 molar basis, the amount for calcium hydroxide required can be represented as 0 .0 .35 multiplied with 74 .09 gram per mole of calcium hydroxide will be equal to 3 .6 .5.
02:22
97 gram per liter further for treatment for thousand liters we have thousand liter 3 .97 grams per liter equals to 3 .97 multiplied 10 raised to power 3 grams which is equal to 3 .97 kilograms.
02:54
We need to add which negative ions so magnesium oxide, sorry, magnesium hydroxide will be mg 2 positive plus 2, which gives, which has already provided the ksp value as 1 .5, multiply 10 -dose to 1 .11.
03:19
When initial concentration of 1 .03, when the initial concentration of 1 .03 5, 5 grams per litre, is reduced to 0 .1 % of the original...