Answer to Question #146678 in Physical Chemistry for Sanaa

Question

Calculate the Ka value of chloroethanoic acid (CH2ClCOOH) given that a solution with a concentration of 0.200 mol dm–3 has a pH of 1.77

4. Which is a stronger acid, Phenol (pKa=9.80) o rethanoic acid(Ka=1.76×10–5).Explain your answer.

Answer 1

3. [CH2ClCOOH] = 0.200M

Ka= ?

pH = 1.77

pH = - log [H+]

1.77 = - log [H+]

 [H+] = 10-1.77

 [H+] = 1.70 x 10-2M

 CH2ClCOOH :::::::::: CH2ClCOO- + H+

The equilibrium concentrations are

CH2ClCOOH = 0.200 - 1.70 x 10-2 = 0.183

CH2ClCOO- = 1.70 x 10-2 

H+ = 1.70 x 10-2

Ka = [CH2ClCOO-][H+]  

      [CH2ClCOOH]

Ka = (1.70 x 10-2)²  

     0.183

Ka = 1.58 x 10-3 moldm-3

4. pKa of phenol is 9.80

  Ka of ethanoic acid is 1.76 x 10-5

To compare these, let's find pKa of ethanoic acid also

pKa = -log Ka

pKa = -log (1.76 x 10-5)

pKa = 4.75

Ethanoic acid is stronger than phenol because the lower the pKa, the higher the strength of an acid. Since ethanoic acid has a lower pKa of 4.75, vis-a-vis the higher pKa of 9.80 of phenol. Ethanoic acid is stronger.