Question

A solution contains 1.577 mg of CoSO4 (155.0 g/mol) per milliliter. Calculate a. the volume of 0.007840 M EDTA needed to titrate a 25.00-mL aliquot of this solution. Volume = mL b. the volume of 0.006901 M Zn2+ needed to titrate the excess reagent after addition of 50.00 mL of 0.007840 M EDTA to a 25.00-mL aliquot of this solution. Volume = mL c. the volume of 0.007840 M EDTA needed to titrate the Zn2+ displaced by Co2+ following addition of an unmeasured excess of ZnY2- to a 25.00- mL aliquot of the CoSO4 solution. The reaction is Co2+ + ZnY2- → CoY2- + Zn2+ Volume =

Answer 1

Solution Contains 1.577 mg of Cosoy per ml. Molar mass = 155 g/mol moles = mass_= 1,577x168 malar mass 155 g mot moles=0.0101742 Moles x10-3 1 ml of solution contains orolol 742 moles i 1000ml a . 10.1742 moles This molarity of Cosoy solution ; INA/00X742 mol/el IM= 0.0101742 M! volume of EDTA needed to titrate 25 ml aliquot; MiK = M₂ V2 Cosoy EDTA 0.0101742 x 14.25 = 0.007940 XV M MLM v=32.4432mL / This, this much volume required (6) Excers reagent added 50mL of 0.007840M EDTA Excess volume = 50-32-4432 mL = 17.56 mL This exceis volume is needed to be titrated with 2, 2+ MV = M₂ V EDTA

0.007840 x 17.56 -0.006001 XV - M m2 M = 19.948 m2 o The reaction taking place; Co²+ + Zny2- Cort Zn2+ We have added any 2- to Roman Solution * 25 ml x 0.0101742 mal = 0.25435 mmol It is clear from equation that; Immol of Co2+ froduces 1 mol of Zn2+ .: 0.25435 mmol 0.25435 mmol qiz > n* displaced by Co2 = 0.25435 monol This much need to be titrated with EDTA This requires 0.25435 mmol of EDTA . EDTA & Z react in lal manner i Volume of EDTA = moles of EDTA Molarity " = 0.25435 mmol 0.007840 moll- IV = 32.4426 mL! Thus, 32.4426 mL of EPTA required to titrate zo ²+.