JEE MainChemistryIonic Equilibrium
A weak monobasic acid HA has a degree of dissociation . A student plots a graph with the pH of the solution on the y-axis and ( 1- ) on the x-axis. What will be the slope and the y-intercept of the resulting straight line?
Options
- ASlope = -1 , Intercept = pK_a
- BSlope = 1 , Intercept = pK_a
- CSlope = 1 , Intercept = -pK_a
- DSlope = -1 , Intercept = -pK_a
Correct answer
B. Slope = 1 , Intercept = pK_a
Step-by-step solution
For a weak monobasic acid HA, the dissociation equilibrium is: HA H ^+ + A ^- The equilibrium constant K_a is given by: K_a = [ H ^+][ A ^-] [ HA ] If the initial concentration is C and the degree of dissociation is , then at equilibrium: [ H ^+] = C [ A ^-] = C [ HA ] = C(1- ) Substituting these into the K_a expression: K_a = (C )(C ) C(1- ) = [ H ^+] 1- Rearranging for [ H ^+] : [ H ^+] = K_a ( 1- ) Taking the negative logarithm on both sides: - [ H ^+] = - K_a - ( 1- ) pH = pK _a + ( 1- ) Comparing this equation