Using 50 cm\(^3\) of 1 M potassium hydroxide and 100 cm\(^3\) of 1 M tetraoxosulphate (VI) acid, calculate the respective volumes in cm\(^3\) of base and acid that would be required to produce the maximum amount of potassium tetraoxosulphate (IV) [K = 39, S = 32, O = 16, H = 1]
First, let's write a balanced equation for the chemical reaction:
2KOH + H\(_2\)SO\(_4\) → K\(_2\)SO\(_4\) + 2H\(_2\)O
Reacting ratio: 2moles : 1 mole 1 mole
Reacting Volume: 50cm\(^3\) : 100cm\(^3\)
Actual volume that reacted: 50cm\(^3\) : 25cm\(^3\)
Residual : 0cm\(^3\) : 75cm\(^3\)
Volume of base (KOH) = 50 cm\(^3\)
Volume of acid (H\(_2\)SO\(_4\)) = 25 cm\(^3\)
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