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?
- 5cm\(^3\)
- 10cm\(^3\)
- 20cm\(^3\)
-
50cm\(^3\)
The correct answer is: D
Explanation
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 = 250.5 = 50cm\(^2\)
Correct computation as it follows the equation specified.
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