Saved Bookmarks
| 1. |
Calculate K_(a) of acetic acif if its 0.05 N solution has equivalent conductance of 7.36 mho cm^(2) at 25^(@)C. "" (Lambda_(CH_(3)COOH)^(@) = 290.7) |
|
Answer» Solution :DEGREE of dissociation (x) `= (Lambda_(c))/(Lambda_(0)) = (7.36)/(390.7) = 0.0188`. For the equilibrium `{:(0.05,0,0,,"Initial concn. (moles/litre)"),(CH_(3)COOH =, CH_(3)COO^(-)+,H^(+),,),(0.05(-1x),0.05 x,0.05x,,"Equilibrium concentration"):}` (for `CH_(3)COOH, 0.05 N = 0.05 M`) `K_(a) = (0.05 x xx 0.05 x)/(0.05(1-x))` Since x is very small, `K_(a) = 0.05 x^(2) = 0.05 xx (0.0188)^(2)` `= 1.76 xx 10^(-5)` mole/litre. |
|