Question

100 ml. of hard water sample require 15 mL. of 0.01 M of EDTA with buffer and EBT indicator. Another 100 mL of the sample is boiled for about half an hour and filtering the ppt., the volume of filtrate is made 100 mL. by the addition of distilled water. 20 mL of this boiled water sample required 5 mL. of 0.001M EDTA following the same procedure. Calculate the total, permanent and temporary hardness of the sample

Answer 1

Answer:

133 is your answer for you questions

Answer 2

Answer:

Total hardness of sample is 150ppm

Permanent hardness of sample is 25ppm

Temporary hardness of sample is 125ppm

Explanation:

Given,

Volume of water sample=100ml

Volume of EDTA=15ml=0.015L

Molarity of EDTA=0.01M

Therefore,no.of moles of EDTA=Molarity x Volume

$= 0.01 \times 0.015 = 15 \times 10 {}^{ - 5} mol$

Hence,

$moles \: of \: Ca {}^{ + 2} = 15 \times 10 {}^{ - 5}$

We know that to represent hardness we require the moles and molar mass of Calcium Carbonate(CaCO3)

$moles \: of \: CaC0_{3} = 15 \times 10 {}^{ - 5} \\ mass = 15 \times 10 {}^{ - 5} \times 100gmol {}^{ - 1} \\ = 0.015gm$

Therefore Total hardness is

$total \: hardness=\frac{mass \: of \: CaC0_{3}}{volume} \times10 {}^{6} \\ = \frac{0.015}{100} \times 10 {}^{6} = 150ppm$

After heating,

Volume of EDTA required=5ml=0.005L

Molarity of EDTA=0.001M

no.of moles of EDTA=Molarity x Volume

$= 0.001 \times 0.005 = 5 \times 10 {}^{ - 6} mol$

Therefore

$moles \: of \: CaC0_{3} = 5 \times 10 {}^{ - 6} \\ = > mass = 5 \times 10 {}^{ - 6} \times 100gmol {}^{ - 1} \\ = 5 \times 10 {}^{ - 4} gm$

Now Permanent hardness is given as

$\frac{mass}{volume \: of \: sample} \times 10 {}^{6} = \frac{5 \times 10 {}^{ - 4} }{20} \times 10 {}^{6} = 25ppm$

Temporary hardness=Total Hardness-Permanent Hardness=150ppm-25ppm=125ppm

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