7.48 Assuming complete dissociation, calculate the pH of the following solutions:

(i) 0.003 M HCl (ii) 0.005 M NaOH (iii) 0.002 M HBr (iv) 0.002 M KOH

(i) 0.003MHCI:H2O+HCIH3O++CI-Since HCI is completely ionized,[H3O+]=[HCI][H3O+]=0.003Now,pH=-log[H3O+]=-log(.003)=2.52Hence, the pH of the solution is 2.52.

(ii) 0.005 M NaOH:NaOH(aq)Na(aq)++HO(aq)-[HO-]=[NaOH][HO-]=0.005pOH=-log[HO-]=-log(0.005)pOH=2.30pH=14-2.30=11.70Hence, the pH of the solution is 11.70.

 

(iii) 0.002 HBr:

HBr+H20H3O++Br-[H3O+]=[HBr][H3O+]=.002pH=-log[H3O+]=-log (0.002)=2.69

Hence, the pH of the solution is 2.69

(iv) 0.002 M KOH:

KOH(aq)K(aq)++OH(aq)-[OH-]=[KOH][OH-]=.002Now, pOH=-log [OH-]=2.69pH=14-2.69=11.31

Hence, the pH of the solution is 11.31