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The pH of 0.004 M hydrazine solution is 9.7 . Calculate its ionization constant K_b and pK_b. |
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Answer» SOLUTION :WEAK base Hydrazine , `NH_2NH_2` the CALCULATION of `[OH^-]` is , pH=9.7 and pH+POH=14 `therefore` pOH=14.0-9.7 = 4.3 `[pOH^-] = -log [OH^-]`=4.3 `therefore log [OH^-]=-4.3 = bar5.7` `therefore [OH^-]`=Antilog `(bar5.7)=5.012xx10^(-5)` =xM Calculation of ionization constant `K_b`: Ionic equilibrium of `(NH_2NH_2)` hydrazine solution is as under. `{:(,NH_2NH_(2(aq))+,H_2O_((l)) hArr, NH_2NH_(3(aq))^(+) + , OH_((aq))^(-)),("At equili." , (0.004-x),M approx x, x M, x M):}` `therefore K_b=([NH_2NH_3^+][OH^-])/([NH_2NH_2])` `=((5.012xx10^(-5))(5.012xx10^(-5)))/0.004` `=6.028xx10^(-7)` Calculation of `pK_b` : `pK_b=-log K_b=-log (6.28xx10^(-7))` =-(0.7980- 7.000) = 6.020 |
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