In the reaction between sodium hydroxide and tetraoxosulphate (VI) solutions, what volume of 0.5 molar sodium hydroxide would exactly neutralize 10cm(^3) of 1.25 molar tetraoxosulphate (vi) acid?
-
A.
25cm(^3) -
B.
10cm(^3) -
C.
20cm(^3) -
D.
50cm(^3)
Correct Answer: Option D
Explanation
Equation of reaction : 2NaOH + H(_2)SO(_4) → Na(_2)SO(_4) + 2H(_2)O
Concentration of a base, CB = 0.5M
Volume of acid, V(_A) = 10cm(^3)
Concentration of an acid, C(_A) = 1.25M
Volume of base, V(_B) = ?
Recall:
(frac{C_A V_A}{C_B V_B} = frac{n_A}{n_B}) … (1)
N.B: From the equation,
(frac{n_A}{n_B} = frac{1}{2})
From (1)
(frac{1.25 times 10}{0.5 times V_B} = frac{1}{2})
(frac{12.5}{0.5V_B} = frac{1}{2})
25 = 0.5V(_B)
VB = 50.0 cm(^3)
There is an explanation video available below.