A) 2
B) 1.3
C) 0
D) 7
Correct Answer: B
Solution :
20 ml. of 0.1 NHCl = \[\frac{0.1}{1000}\times 20g\] eq. = \[2\times {{10}^{-3}}g\] eq. 20ml. of 0.001 KOH = \[\frac{0.001}{1000}\times 20\,gm\]eq. = 2 × \[{{10}^{-5}}g\]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. \ [HCl] = \[\frac{1.98\times {{10}^{-3}}}{40}\times 1000M\] = \[4.95\times {{10}^{-2}}\] \[\therefore \] pH = \[2-\] log 4.95 = 2 ? 0.7 = 1.3.
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