Question

In a water Hardness complexometric micro-titration, EDTA was used as a standard. The formula is : Ca ^2+ Y^4- - CaY^2 where (Y is EDTA) Suppose the EDTA used was not as dry as it should have been a. Would [EDTA] in the EDTA solution be higher or lower than the actual value? b. Would the [Ca] determined in the unknown be higher or lower than the actual value?

Answer 1

a) Due to presence of water, the actual mass of EDTA taken will be less than required mass. Hence, [EDTA] concentration in the EDTA solution will be lower than the actual value. For example, suppose we take 100 g of EDTA out of which 20 g are water and 80 g is EDTA. Suppose M be the molar mass of EDTA and total volume of EDTA solution be 1 L. Hence, $[EDTA] = \dfrac {80}{M \times 1} \: mol/L$

But if we take 100 g of dry EDTA, then, the actual

$[EDTA] = \dfrac {100}{M \times 1} \: mol/L$

b) The [Ca] determined in the unknown will be higher than the actual value.

Thus in above example, i we use 100 g of wet EDTA and if we assume that the volume of Ca solution is also 1 L, then $[Ca] = \dfrac {80}{M \times 1} \: mol/L$ This the actual value. But if we use wet EDTA then we will determine $[Ca] = \dfrac {100}{M \times 1} \: mol/L$