In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5M sodium hydroxide would exactly neutralise 10cm\(^3\) of 1.25M sulphuric acid?
\(2NaOH + H_{2}SO_{4} = Na_{2}SO_{4} + 2H_{2}O\)
V x 0.5M = 2(10 x 1.25)
0.5V = 2 x 12.5
0.5V = 25
V = \(\frac{25}{0.5}\) = \(50cm^{3}\)
Alternatively, \(\frac{{C_A}{V_A}}{{C_B}{V_B}}\) = \(\frac{N_A}{N_B}\)
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