A) 0.25 N basic
B) 0.2 N acidic
C) 0.25 N acidic
D) 0.2 N basic
Correct Answer: D
Solution :
\[{{V}_{1}}=20\,\,ml,\,\,{{N}_{1}}=0.25N,\,\,{{V}_{2}}=30\,\,ml\] \[{{N}_{2}}=0.2N\] \[\therefore N=\frac{{{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}=\frac{20(0.25)+30(0.2)}{20+30}\] \[=\frac{5+6}{50}=\frac{11}{50}=0.2N\,\]basic
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