Determination of SO4 in drinking water by using [EDTA]-2

Determination of So₄^-2 in drinking water by using [EDTA]^-2

Principle:

The permissible limit of So₄^-2 in drinking water is 200mg/L and its excess can cause severe problems related to the liver and stomach. For the determination of So₄^-2 content in drinking water volumetric method of analysis using [EDTA]^-2 is an indirect way.

The complex which is formed between [EDTA]^-2 and metal is always colourless with almost every type of titration involving [EDTA]^-2. We need a buffer solution having pH=10 because maximum complex formation occurs at this pH. Moreover [EDTA]^-2 has six binding sites I.e. ( Hexa dentate ) and it is stable in the form of disodium which replaces its two negative charges.

BaCl₂ + [EDTA]^-2 --> Ba[EDTA]^-2

The indicator used in this experiment is EBT which is Eriochrome black T, sometimes also known as solo chrome black T.

The Colour of this complex is red while the colour of the above-formed Ba[EDTA]^-2 complex is always colourless. On titration, [EDTA]^-2 replaces the EBT and at the end of the titration when all of the EBT is replaced by the [EDTA]^-2, we have free EBT with having blue colour. i.e

Apparatus and chemicals:

Beaker 50mL, 100mL, stand , burette , conical flask , 0.01M [EDTA]^-2 , 0.01M BaCl₂ , EBT, buffer solution etc..

Chemical equation:

Procedure:

1)      10mL of the sample containing So₄^-2 content has been taken in a conical flask.

2)      Followed by 15mL of the BaCl₂ with 5mL of the buffer solution and 2 to 3 drops of EBT.

3)      The initial colour was red which on titration against standard [EDTA]^-2 converted into blue colour.

4)      The same procedure has been repeated three times and its mean volume was noted. 

Observation and calculations:

Sr no	Initial volume	Final volume	volume of [EDTA]-2 used
01	0.0mL	8.5Ml	8.5mL
02	9.0mL	17.5mL	8.5mL
03	18mL	26.6mL	8.6mL

                                                            Mean   =8.5 + 8.5 + 8.6 /3 = 8.5mL

1000mL of BaCl₂ has no of moles = 0.01 mol

1mL of BaCl₂ has no moles            = 0.01/1000

15mL of BaCl₂ has no of moles     = 0.01/1000x15 = 0.00015 mol

Now

1000mL of [EDTA]^-2 has no of moles = 0.01 mol

1mL of   [EDTA]^-2 has no moles          = 0.01/1000

8.5mL of [EDTA]^-2 has no moles        = 0.01/1000x8.5 = 0.000085 mol

Now

Unconsumed moles of BaCl₂ = Total moles of Bacl₂ – moles of BaCl₂ with [EDTA]^-2

                                                 = 0.00015    -       0.000085

                                                 = 0.000065mol

From the equation,

One mole of Ba   = one mole of SO₄^-2

0.000065              = 0.000065

So,

10mL of the sample has no of moles of SO₄^-2 = 0.000065 mol

1mL of the sample has no of moles of SO₄ ^-2  = 0.000065/10

1000mL of the sample has no of moles of SO₄^-2 = 0.000065/10 X 1000 = 0.0065mol

So,

Strength in g/L = molarity X molecular weight

                        = 0.0065    X 96

                        = 0.624 g/L

So,

1000ml has no of moles of SO4 = 0.624

1mL has no of moles of SO₄^-2     = 0.624/1000

500mL has no of moles of SO₄^-2 = 0.624/1000 x 500 =0.312g

Now,

Percentage error = 0.312/1.00 X 100 = 31.2

Now

Percentage error = 1-0.312 = 68.8%

Result:

Hence the given water sample contains 0.624g of SO₄^-2 particles.