In order to standardize a KMnO4 solution, 0.2848 g Fe(NH4)2(SO4)2·6H2O was dissolved in 25 mL 0.18 M H2SO4. The KMnO4 solution was added to the Fe(NH4)2(SO4)2·6H2O solution until a pale pink color persisted. The titration took 24.2 mL of KMnO4 solution. What is the concentration of the KMnO4 solution?
2 answers
You have to use the M1V1=M2V2 equation for this problem and plug in the values into the appropriate parts of the equation and solve for the missing molar concentration.
No, no, Bridget. That will not work in this case BECAUSE 1 mol Fe is not = 1 mol MnO4^-.
1. Write and balance the equation.
Fe(NH4)2(SO4)2.6H2O + KMnO4 + H2SO4 ==> Fe2(SO4)3 + (NH4)2SO4 + K2SO4 + MnSO4 + H2O
2. Convert 0.2848g Fe(NH4)2(SO4)2.6H2O to mols. mol = grams/molar mass = ?
3. Using the coefficients in the balanced equation, convert mols Fe(NH4)2(SO4)2.6H2O to mols KMnO4.
4. Then M KMnO4 = mols KMnO4/L KMnO4. You know mols and L, solve for M.
1. Write and balance the equation.
Fe(NH4)2(SO4)2.6H2O + KMnO4 + H2SO4 ==> Fe2(SO4)3 + (NH4)2SO4 + K2SO4 + MnSO4 + H2O
2. Convert 0.2848g Fe(NH4)2(SO4)2.6H2O to mols. mol = grams/molar mass = ?
3. Using the coefficients in the balanced equation, convert mols Fe(NH4)2(SO4)2.6H2O to mols KMnO4.
4. Then M KMnO4 = mols KMnO4/L KMnO4. You know mols and L, solve for M.
5 answers