The result is 4.16 L.
Based on the provided information, we calculated the following:
Molarity = 0.225 M,
Quantity of KI = 0.935 moles,
To find Volume: Molarity = moles/Volume, hence Volume = moles/Molarity.
Thus, Volume = 0.935/0.225, giving Volume = 4.16 L.
Consequently, 4.16 L of KI is required.
The neutralization reaction that occurs between potassium hydroxide and sulfuric acid can be represented as follows
2KOH + H2SO4 ---> K2SO4 + 2H2O
The quantity of moles of KOH is calculated using (43.74 x 0.500)/ 1000 = 0.02187 moles
Given that the stoichiometric ratio of KOH to H2SO4 is 2:1, the moles of H2SO4 can be determined as 0.02187/2 = 0.01094 moles
To find the concentration (molarity), use the formula (0.01094/50) x 1000 = 0.2188M