Answer to Question #109167 in General Chemistry for Michael

Question

0.0089M KMnO4 was titrated against 25cm3 of a mixture of Fe(II) and Fe(III) salts acidified with 20cm3 of 1M dil H2SO4, 22.20cm3 of KMnO4 solution required. Calculate
1. Concentration of Fe(II) in moldm-3 and gdm3 In mixture
2. Ratioof Fe(II) to Fe(III) in mixture of it contains 9.80gdm-3
3.percentage purity of Fe(III) salt in mixture

Answer 1

$5Fe^2+MnO_4^-+8H^+=5Fe^3+Mn^2+4H_2O$

the reaction at titration

$\nu(MnO_4^-)=\nu(KMnO_4)=V(KMnO_4)*C(KMnO_4)$

$\nu(MnO_4^-)<<\nu(H^+)$

$\nu(MnO_4^-)=22,2 cm^3 * 0,0089 Moles/1000 cm^3 =0,00019758mol$

$5*\nu(MnO_4^-)=\nu(Fe^2+)=0,0009879moles$

$C(Fe^2+)=\nu(Fe^2+)/Voverall$

$Voverall = V(KMnO_4)+V(Fe^2//Fe^3) + V(H_2SO_4)$

Voverall==22,2cm³+25cm³+20cm³=67,2 cm³

1. C(Fe²⁺)=0,0147moles/dm³

1*. w(Fe²⁺)=C(Fe²⁺)*Molar/Dm³=0,8210g/dm³

w(Fe²⁺)/w(Fe³⁺)=0,821/(9,8-0,821)=0,09144

No specific salt anion is indicated (for example, it could be a complex like EDTA, which always binds to a metal ion in a 1 to 1 ratio).

Therefore, we will calculate based on the mass fractions calculated earlier.

(However, a calculation through equivalents would perhaps be more correct.)

(9,8-0,821)/9,8 * 100% = 91,62%