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The indicator constant of phenolphthalein is approximately `10^(-8)`.A solution is prepared by adding 100.01 c.c. of 0..01 N sodium hydroxide to 100.00 c.c. of 0.01 N hydrochloric acid.If a few drops of phenolphthalein are now added.what traction of the indicator is converted to its coloured form ?A. `1/3`B. `3/4`C. `1/2`D. `9/11` |
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Answer» Correct Answer - A millimole, `underset(100.01xx0.01)(NaOH) underset(" " 100xx0.01)(+HCl)toNaCl+H_2O` So, NaOH left =0.0001 Hence, concentration of `OH^(-) =0.0001/200.01=0.0001/200` `HPhhArr H^+ + Ph^(-)` `K_a=([H^+][Ph^(-)])/([HPh])` `implies 10^(-8)=(k_w[Ph])/([OH^(-)][HPh])=((10xx^(-14)xx[Ph])/0.0001)/(200)[HPh], " " ([Ph^(-)])/([HPh])=1/2` so `([Ph^(-)])/([Ph^(-)]+[HPh])=1/3` |
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