0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution. Th e pH of the resulting solution would be

0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution

1.	0.365 g of HCl gas was passed through 100 cm^(3) of 0.2 MNaOH solution. Th e pH of the resulting solution would be
Answer» 1 5 8 13 Solution :`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`