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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 

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(i) 0.003MHCl: 

Since HCl is completely ionized, 

Now, 

Hence, the pH of the solution is 2.52.

(ii) 0.005MNaOH: 

Hence, the pH of the solution is 11.70. 

(iii) 0.002 HBr: 

Hence, the pH of the solution is 2.69. 

(iv) 0.002 M KOH: 

Hence, the pH of the solution is 11.31. 

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