PRACTICAL CHEMISTRY
37509
What is the molarity of \({H_2}S{O_4}\) solution if \(25\, ml\) is exactly neutralised with \(32.63\, ml\) of \(0.164 \,M\), \(NaOH\)
1 \(0.107 \,M\)
2 \(0.126 \,M\)
3 \(0.214 \,M\)
4 \(-0.428 \,M\)
Explanation:
(a) \( 0.164\, M\, NaOH\ ,\cong \,0.164\,N\,NaOH\)
We know, \({N_1}{V_1} = {N_2}{V_2};\) \({N_1} \times 25 = 0.164 \times 32.63\)
\(0.214\;N\;{H_2}S{O_4} \cong \frac{{0.214}}{2}M\;{H_2}S{O_4}\) (basicity of \({H_2}S{O_4}\;is\;2\))
\( \cong \;0.107\;M\;{H_2}S{O_4}\)