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0.365 g of HCl gas was passed through `100 cm^(3)` of 0.2 M NaOH solution. Th e pH of the resulting solution would beA. 1B. 5C. 8D. 13 |
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Answer» Correct Answer - D `0.365 g HCl =(0.365)/(36.5) "mole" = 0.01 "mole"` `100 cm^(3) "of" 0.2 M NaOH=(0.2)/(1000)xx100` `=0.02 ` mole NaOH left unneutralized = 0.01 mole Volume of solution = 100 ml `:.` Molarity of NaOH in the solution `=(0.01)/(100)xx1000=0.1M = 10^(-1)M` `:. [H^(+)]=(10^(-14))/(10^(-1)M)=10^(-13)M :. pH=13` |
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