Question

A weak monobasic acid is 12% dissociated in 0.15 M solution. What is percent dissociation in 0.30 M solution.​

Answer 1

Answer:

Explanation:

                                  HA-------------> H⁺  +   A⁻

at time, t=0                c                      0        0

at time , t=T             c-cα                   cα       cα

                         

                    Κ     =         [cα][cα] / [c-cα]

                           = cα² / 1-α

here percentage dissociation = 12% , hence degree of dissociation α = 0.12

   1 -  α  is nearly equal to 1 therefore that term can be neglected from the denominator                  

so,   Κ = cα²

       K = 0.15 x 0.12 x 0.12 = 2.1 x 10⁻³

K always remains constant regardless of concentration or degree of dissociation

          NOW CALCULATING DEGREE OF DISSOCIATION αₓ AT 0.3 M                                         CONCENTRATION  

αₓ = $\sqrt{\frac{K}{c} }$

    =$\sqrt{\frac{2.1 * 10^-^3}{0.3} }$

αₓ  = 0.836

PERCENTAGE DISSOCIATION = 83.6%