The approximate pH of a solution formed by mixing equal volumes of solutions of 0.1 M sodium propionate and 0.1 M propanoic acid (the dissociation constant of propanoic acid is 1.3 × 10-5 mol dm -3 ) will be

Question

The approximate pH of a solution formed by mixing equal volumes of solutions of 0.1 M sodium propionate and 0.1 M propanoic acid (the dissociation constant of propanoic acid is 1.3 × 10-5 mol dm -3 ) will be (A) 2.45 (B) 4.89 (C) 5.98 (D) 6.89

Tardigrade Admin · Accepted Answer

As CH 3 CH 2 COONa is a strong electrolyte.

Now consider solution of 0.1 M propanoic acid having initially 0.1 M CH 3 CH 2 COO -.

Ka =([ CH 3 CH 2 COO -( aq )][ H +]/[ CH 3 CH 2 COOH ])=1.3 × 10-5 ((0.1+ X )( X )/(0.1- X ))=1.3 × 10-5 Assuming that 0.1+X=0.1 0.1-X=0.1, we get ((0.1)( X )/(0.1))=1.3 × 10-5 or X =1.3 × 10-5=[ H +] So pH =- log (1.3 × 10-5)=4.89

Q. The approximate pH of a solution formed by mixing equal volumes of solutions of 0.1M sodium propionate and 0.1M propanoic acid (the dissociation constant of propanoic acid is 1.3×10−5moldm−3 ) will be

2.45

4.89

5.98

6.89

Q. The approximate $p H$ of a solution formed by mixing equal volumes of solutions of $0.1 M$ sodium propionate and $0.1 M$ propanoic acid (the dissociation constant of propanoic acid is $1.3 \times 1 0^{- 5} m o l d m^{- 3}$ ) will be