Equilibrium



  • 0_1551522840646__20190302_160336.jpg
    Ans is 2



  • @yogita-panchal

    First we should write the reaction,
    For the HCl...............> H+........+..Cl-
    0.1M.................0................0
    0........................0.1M.........0.1M

    after its complete dissociation
    [Here ahve neglected 10-7M of H+ due to water dissociation because 0.1M is more than 100 times than 10-7M]
    Now for the weak acid which not dissociate completely and always form an equilibrium:
    CH3COOH..+ H2O...=......CH3COO-...+....H+
    0.01M..............................0.................0.1M(from dissociation of HCl) ......initially
    0.01-x..................................x..........x+0.1M.........after dissociation of acetic acid
    Now writing
    Ka=[CH3COO-][H+]/[CH3COOH]
    1.810-5 = x(0.1+x)/0.01
    Approximation:the acid is weak and so assuming

    x<<0.1M,0.1+x~0.1 M

    0.1x= 1.80.0110-5 M

    so x= 1.8*10-6 M
    so % onisation of acid=x/0.01 100=1.810-2%=0.018%(very less due to common ion effect)


Log in to reply
 

Powered by dubbtr | @2019

Looks like your connection to dubbtr was lost, please wait while we try to reconnect.