The concentration (molarity) of [H₃O⁺]=0.776 M
Further explanation
We balance the reaction first so that the number of atoms and the charge of the product and reactant are the same
Reaction(balanced) :
H₂S⇒2H⁺+S²⁻ (eq 1)
2H⁺+6H₂O⇒4H₃O⁺+2OH⁻ (eq 2)
mass of H₂S=0.33 g
mol H₂S (MW=34 g/mol) :
![\tt mol=(0.33)/(34)=0.0097](https://img.qammunity.org/2021/formulas/chemistry/high-school/hz05m4phdfuwso9s06mvyo4en2kona7ttq.png)
From equation 1, mol H⁺ = 2 x mol H₂S , so mol H⁺ =
![\tt 2* 0.0097=0.0194](https://img.qammunity.org/2021/formulas/chemistry/high-school/lbg2dcii1gzba3tdlen967a2wm8gx9b4vt.png)
From equation 2, mol H₃O⁺ = 2 x mol H⁺(4 : 2) , so mol H₃O⁺ :
![\tt 2* 0.0194=0.0388](https://img.qammunity.org/2021/formulas/chemistry/high-school/33wj58slpkihp0fkezi8ctp87qbx7fj8h4.png)
Or simply mol H₃O⁺ = 4 x mol H₂S = 4 x 0.0097 = 0.0388
the molarity of [H₃O⁺] =
![\tt M=(n)/(V)=(0.0388)/(0.05~L(50~ml))=0.776](https://img.qammunity.org/2021/formulas/chemistry/high-school/ypf1w0m421mghjw35kc3errti12abrr4ax.png)