A) 2
B) 1.3
C) 0
D) 7
Correct Answer: B
Solution :
\[\text{20 mL}\] of \[0.1\text{ }N\text{ }HCl=\frac{0.1}{1000}\times 20\,g\,eq\] \[=2\times {{10}^{-3}}g\,eq\] \[\text{20 mL}\]of \[0.001\,KOH=\frac{0.001}{1000}\times 20\,g\,eq\] \[=2\times {{10}^{-5}}g\,eq\] \[\therefore \] HCl left unneutralised \[=2({{10}^{-3}}-{{10}^{-5}})\] \[=2\times {{10}^{-3}}(1-0.01)\] \[=2\times 0.99\times {{10}^{-3}}=1.98\times {{10}^{-3}}g\,eq.\] Volume of solution = 40 mL \[\therefore \] \[[HCl]=\frac{1.98\times {{10}^{-3}}}{40}\times 1000\,\text{M}\] \[=4.95\times {{10}^{-2}}\] \[\therefore \] \[pH=2-\log 4.95\] \[=2-0.7=1.3\]
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