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  • Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl

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

(a)      0.003 M HCl

(b)      0.005 M NaOH

(c)      0.002 M HBr 

 (d)       0.002 M KOH

Answers (1)

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Assuming the complete dissociation. So, \alpha =1

(a) The ionisation of hydrochloric acid is 
HCl\rightleftharpoons H^++Cl^-
Since it is fully ionised then [H^+]=[Cl^-]=0.003M
Therefore, 
pH of the solution = -\log (0.003)
                                = 3-\log (3)
                                = 2.52 

(b) The ionisation of 0.005 M NaOH
  NaOH\rightleftharpoons Na^++OH^-
[Na^+]=[OH^-]=0.005M
Therefore,
pOH of the solution = 
                                   \\=-\log(0.005)\\ =3-\log5\\ =2.301
pH of the solution is equal to (14 - 2.301 =11.70) 

(c)  The ionisation of 0.002 M HBr
  HBr\rightleftharpoons H^++Br^-
[Br^-]=[H^+]=0.002M
Therefore,
pH of the solution = 
                                   \\=-\log(0.002)\\ =3-\log2\\ =2.69
pH of the solution is equal to (2.69) 

 

(d)  The ionisation of 0.002 M\ KOH
  KOH\rightleftharpoons K^++OH^-
[OH^-]=[K^+]=0.002M
Therefore,
pOH of the solution = 
                                   \\=-\log(0.002)\\ =3-\log2\\ =2.69
pH of the solution is equal to (14 - 2.69 = 11.31) 

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manish

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