$10 \;mL$ iron(II) sulphate solution of unknown molarity are titrated with a $0.02\;M $ potassium permanganate solution. The equivalence point is reached after adding $9.4\;mL$ of the permanganate solution. Remember → this reaction and give its ionic and "molecular" equation!
$2K^+$ $+$ $2MnO_4^-$ $+$ $10Fe^{2+}$ $+$ $10SO_4^{2-}$ $+$ $16H^+$ $+$ $8SO_4^{2-}$ $\longrightarrow$ $2Mn^{2+}$ $+$ $2SO_4^{2-}$ $+$ $10Fe^{3+}$ $+$ $15SO_4^{2-}$ $+$ $2K^+$ $+$ $SO_4^{2-}$ $+$ $8H_2O$ $2KMnO_4$ $+$ $10FeSO_4$ $+$ $8H_2SO_4$ $\longrightarrow$ $2MnSO_4$ $+$ $5Fe_2(SO_4)_3$ $+$ $K_2SO_4$ $+$ $8H_2O$ Potassium permanganate and iron(II) sulfate and sulfuric acid produce manganese(II) sulfate and iron(III) sulfate and potassium sulfate and water.
What is the ratio by which iron sulfate and potassium permanganate react ?
Moles of iron sulfate/ moles of potassium permanganate = $\frac{10}{2}$ $=$ $\frac{5}{1}$
Calculate the number of moles of permanganate used!
$n_{KMnO_4}$ $=$ $9.4\cdot 10^{-3}\cdot 0.02$ $=$ $1.88\cdot 10^{-4}\; mol$
Calculate the number of moles of iron sulfate initially present!
$n_{FeSO_4}$ $=$ $5\cdot 1.88\cdot10^{-4}= 9.4 \cdot 10^{-4}\; mol$
Calculate the molarity of the initial iron sulfate solution!
$c_{FeSO_4}$ $=$ $\frac{9,4 \cdot10^{-4}}{10\cdot10^{-3}}$ $=$ $0.094\; M$