A) 1
B) 5
C) \[1.3\]
D) 9
Correct Answer: C
Solution :
\[20\,\,ml\] of \[0.1\,\,NHCl\] \[=\frac{0.1}{1000}\times 20\,g.eq.=2\times {{10}^{-3}}g\,eq.\] \[20\,\,ml\] of \[0.001\,\,NaOH\] \[=\frac{0.001}{1000}\times 20\,g.eq=2\times {{10}^{5}}\,g\,eq.\] \[\therefore \] \[HCl\] left unneutrialised \[=2({{10}^{-3}}-{{10}^{-5}})\] \[=2\times {{10}^{-3}}(1-0.01)\] \[=2\times 0.99\times {{10}^{-3}}=1.98\times {{10}^{-3}}\,g.eq.\] Volume of solution\[=40\text{ }ml\] \[[HCl]=\frac{1.98\times {{10}^{-3}}}{40}\times 1000M=4.95\times {{10}^{-2}}\] \[pH=-\log \left[ 4.95\times {{10}^{-2}} \right]=2-\log 4.95=2-0.7=1.3\]
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