Question

select all that apply
an aqueous solution of hydrogen peroxide is used in hospitals to prevent the transmission of infections. hydrogen peroxide reacts with permanganate ion in solution according to the balanced equation given below.
5h₂o₂(aq) + 2mno₄⁻(aq) + 6h⁺(aq) → 5o₂(g) + 2mn²⁺(aq) + 8h₂o(l)
a 20.00 - ml sample of hydrogen peroxide solution was titrated with acidified kmno₄ solution. 47.5 ml of 0.105 m kmno₄ solution was required to reach the equivalence point. which of the following options correctly reflect the steps required to calculate the concentration of h₂o₂?
the concentration of the peroxide solution is 0.625 m
the concentration of the peroxide solution is 0.263 m.
there are 2.00 × 10⁻³ moles of h₂o₂ in the sample.
4.99 × 10⁻³ moles of mno₄⁻ are reacted

Explanation:

Step1: Calculate moles of \(MnO_4^-\)

Use the formula \(n = C\times V\) (where \(n\) is moles, \(C\) is concentration, \(V\) is volume in liters).
\(V = 47.5\space mL=47.5\times10^{- 3}\space L\), \(C = 0.105\space M\)
\(n(MnO_4^-)=0.105\space mol/L\times47.5\times 10^{-3}\space L = 4.99\times10^{-3}\space mol\)

Step2: Use mole - ratio from the balanced equation

From \(5H_2O_2(aq)+2MnO_4^-(aq)+6H^+(aq)\to5O_2(g)+2Mn^{2 +}(aq)+8H_2O(l)\), the mole - ratio \(n(H_2O_2):n(MnO_4^-)=5:2\)
\(n(H_2O_2)=\frac{5}{2}\times n(MnO_4^-)\)
\(n(H_2O_2)=\frac{5}{2}\times4.99\times 10^{-3}\space mol=1.25\times10^{-2}\space mol\)

Step3: Calculate concentration of \(H_2O_2\)

\(V(H_2O_2)=20.00\space mL = 20.00\times10^{-3}\space L\)
\(C(H_2O_2)=\frac{n(H_2O_2)}{V(H_2O_2)}=\frac{1.25\times10^{-2}\space mol}{20.00\times10^{-3}\space L}=0.625\space M\)

Answer:

The concentration of the peroxide solution is \(0.625\space M\) (first option), \(4.99\times 10^{-3}\space moles\space of\space MnO_4^-\) are reacted (fourth option)