Titration reaction: HY_3-(aq) + Ca_2 + (aq) rightarrow CaY_2 - (aq) + H + (aq) (

Question

Titration reaction: HY_3-(aq) + Ca_2 + (aq) rightarrow CaY_2 - (aq) + H + (aq) (also for Mg_2+) End point reaction: HY_3-(aq) + MgIn-(aq) rightarrow MgY_2-(aq) + HIn_2(aq) Since the indicator requires a trace of Mg_2+to operate properly, a little magnesium ion will be added to each solution. The effect of the added Mg_2+ can be subtracted by titrating a blank. A 0.2431 g sample of CaCO3 is dissolved in 6 M HCI and the resulting solution is diluted to 250.0 mL in a volumetric flask. Titration of a 25.00 mL sample of the solution requires 28.55 mL of EDTA to reach the Eriochrome Black T end point. A blank containing the same amount of Mg_2+ requires 2.60 mL of EDTA. What is the molarity of the EDTA solution?

Explanation / Answer

Given mass of CaCO3 then calculate  moles CaCO3

CaCO3 MW = 100.087 g/mole

mole of CaCO3 = mass of CaCO3 / molar mass of CaCO3

0.2431 g CaCO3 = 0.2431 g / 100.087 g/mole = 0.002429 moles CaCO3

The CaCO3 was diluted to250 ml i.e. 0.250 L :

0.02429 mole/0.250l = 0.009716 M solution of CaCO3

then a 25 ml aliquot was taken

25 ml * 0.009716 moles/1000ml = 0.000243 moles CaCO3 in aliquot

so here blank =2.60ml & sample reading=28.55ml therefore

titration reading= sample-Blank i.e. 28.55-2.60=25.95ml

0.000243 moles CaCO3 in aliquot This was titrated with 25.95 ml EDTA

25.95ml * M EDTA/1000 ml = 0.000243 moles

M EDTA = 0.009364 M

so molarity of EDTA=0.0094 M or 9.4mM

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