A) 2.5
B) 1.5
C) 2
D) 1
Correct Answer: C
Solution :
For \[\text{HCl}\]molarity is equal to normality. So, the normality of the resulting mixture is calculated as \[\text{N}\,\text{= }\frac{{{N}_{1}}{{V}_{1}}+N{{}_{2}}{{V}_{2}}}{{{V}_{1}}+{{V}_{2}}}\] \[=\frac{0.015\times 100+0.005\times 100}{100+100}\] \[=\frac{1.5+0.5}{200}=\frac{2}{200}=\frac{1}{100}={{10}^{-2}}\] Normality of resulting mixture \[={{10}^{-2}}N\] Resulting solution is acidic in nature. So, this resulting normality represent the \[[{{H}^{+}}]\]concentration. Then, \[[{{H}^{+}}]={{10}^{-2}}\] \[pH=-\log [{{H}^{+}}]\] \[=\log \frac{1}{[{{H}^{+}}]}=\log \frac{1}{{{10}^{-2}}}\] \[=\log {{10}^{2}}\] \[=2\log 10\] \[=2\]
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