0.28 g of CaCO3 was dissolved in 1 liter of water. 100 ml of this water on titration required 28 mL of EDTA. 100 mL of unknown hard water sample required 33 mL of same EDTA. After boiling and cooling 100 mL of this sample required 10 mL of EDTA. Calculate the temporary and permanent hardness
Answers
Answer:
Explanation:
Meq. of CaCO
3
dissolved =
2
100
10
×1000=200
Meq. of HCl added =250×1=250
∴ Meq. of HCl after reaction with CaCO
3
=250−200=50
Now, meq. of HCl left = meq. of KOH used
⟹50=2×V ⟹V=25 mL
Hence, volume required is 25 mL.
Answer:
230 ppm or mg/l
Explanation:
step 1 - strength of EDTA solution given 1 liter of standard
Hance, each ml of s. contains 0.28 caco3
AS,
28ml of EDTA = 100ml of hardwater
= 100×0.28 mg of caco3
= 28mg of caco3
1 ml of EDTA = 28/28 mg cacao3
= 1 mg of caco3
step 2 - Determination of hardness of water
Given,
100ml of unknown hardwater sample = 33 ml of EDTA
= 33×1
= 33 mg og caco3
1000 ml (1 L ) of unknown hardwater sample
= 33×1000/100 mg of caco3
Total hardness = 330 ppm or mg/l
100ml of boiled water sample = 10 ml of EDTA
= 10×1 mg of caco3
= 10 mg og caco3
1000ml (1L) of boiled water = 10×1000/100 mg of caco3
= 100mg/l or ppm
Temporary hardness = Total hardness - permanent hardness
= 330 - 100
= 230 ppm or mg/l