Question

What is [h+] in mol/l of a solution that is 0.20 m ch3coona and 0.10 m in ch3cooh ?

Answer 1

Answer:9.0*10^-6 mol/L

Explanation:

The ionisation of CH3 COOH and CH3 COONa takes places as follows :-

CH3 COONa→CH3 COO−  + Na+

CH3 COOH⇌CH3 COO−  + H+

Due to the presence of common CH3 COO−  ions, the equilibrium of the acid will shift towards left (by Le Chateliers's principle)

As we know, Ka  =  [CH3 COO− ][H+ ]

                            ---------------------------

                                 [CH3COOH]          (this is an equation)

Here the [CH3COO− ]=0.2M and [CH3 COOH]=0.1M and Ka =1.8×10^−5

and hence substituting the values in the above formula, we can get the value of concentration of [H+ ] ions.

∴ 1.8×10^−5   =        [0.2][H+ ]

                        ----------------------

                                    [0.1]       (this is an equation)​  

∴[H+]=9.0×10−6  mol/L