JAMB - Chemistry (1978 - No. 4)
In the reaction between sodium hydroxide and sulphuric acid solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralise 10cm3 of 1.25 molar sulphuric acid?
5cm3
10cm3
20cm3
25cm3
50cm3
Wyjaśnienie
\(2NaOH + H_{2}SO_{4} = Na_{2}SO_{4} + 2H_{2}O\)
V x 0.5M = 2(10 x 1.25)
0.5V = 2 x 12.5
0.5V = 25
V = \(\frac{25}{0.5}\) = \(50cm^{3}\)
V x 0.5M = 2(10 x 1.25)
0.5V = 2 x 12.5
0.5V = 25
V = \(\frac{25}{0.5}\) = \(50cm^{3}\)