Question

Calculate the volume of 8.87×10-2 M calcium hydroxide required to neutralize 15.0 mL of a 0.389 M hydrobromic acid solution. mL

Answer 1

Answer : The volume of calcium hydroxide is, 32.89 ml

Explanation :

Using neutralization law,

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1$ = basicity of an acid = 1

$n_2$ = acidity of a base = 2

$M_1$ = concentration of hydrobromic acid = 0.389 M

$M_2$ = concentration of calcium hydroxide = 0.0887 M

$V_1$ = volume of hydrobromic acid = 15 ml

$V_2$ = volume of calcium hydroxide = ?

Now put all the given values in the above law, we get the volume of calcium hydroxide.

$1\times 0.389M\times 15ml=2\times 0.0887M\times V_2$

$V_2=32.89ml$

Therefore, the volume of calcium hydroxide is, 32.89 ml