Question
$$100\,c{m^3}$$ of $$0.1\,NHCl$$ is mixed with $$100\,c{m^3}$$ of $$0.2\,N\,NaOH$$ solution. The resulting solution is
A.
$$0.1\,N$$ and the solution is basic
B.
$$0.1\,N$$ and the solution is acidic
C.
$$0.05\,N$$ and the solution is basic
D.
$$0.05\,N$$ and the solution is acidic
Answer :
$$0.05\,N$$ and the solution is basic
Solution :
$$\eqalign{
& {\text{Normality}} \cr
& = \frac{{{N_1}{V_1} - {N_2}{V_2}}}{{{V_1} + {V_2}}} \cr
& = \frac{{0.2 \times 100 - 0.1 \times 100}}{{100 + 100}} \cr
& = \frac{{10}}{{200}} \cr
& = 0.05N\,NaOH \cr} $$