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1 mL of 10−5 M HCl was diluted to 1 L by adding water. The pH of the resultant solution is A. 8. B. 6.9. C. 5. D. 7.1.

 1 mL of 10−5 M HCl was diluted to 1 L by adding water. The pH of the resultant solution is
A. 8. 
B. 6.9.
C. 5.
D. 7.1.


1 Answers

25763 Points
2 years ago

You know that water autoionises to H+ and OH-.

2 H2O

= 8 - 1.02 =6.98

pH = - log [H+] = - log 10.5*10^-8 = 8 - log10.5

(Note that x > C, more than 90% of H+ in solution is due to autodissociation of water and not due to the acid.)

[H+] = C+x = 10^-8 + 9.5*10^-8 = 10.5*10^-8

x = 9.5*10^-8

Now, Kw= 10^-14 at 25 degree Celsius and if C = 10^-8

If Kw

x ={ - C +√(C^2+ 4*Kw)}/2

(C+x)*x = Kw => x^2 + C*x - Kw =0

Now, [H+]*[OH-] = Kw

Let the molarity of the monoprotic strong acid be C, then [H+] = C +x , where x is the contribution from autodissociation of water.[OH-] = x, because autodissociation of water gives equal amount of H+ and OH-.

If the molarity of the acid solution is 10^-6 M or lower [H+] from auto dissociation of water must be taken into account. In such cases we can proceed as follows.

When 10^-5 M HCl is diluted 1000 times the concentration is reduced to 10^-8 M. If you don't consider the contribution of water towards [H+] , you get pH= 8, which is ridiculous since an acid solution can not be alkaline.

The product of the concentrations [H3O+]*[OH-] is a constant that is called the ionic product of water ( Kw). In every aqueous solution [H+]*[OH-] = Kw.. If [H+] rises then there must be a fall in [OH-] to satisfy the above condition. In an aqueous solution of an acid , unless its concentration is very low, the acid is the main source of H+ ion , the contribution from autodissociation of water being negligible due to common ion effect.( Common ion effect is nothing but the Le Chatelier principle applied to ionic solutions).When the acid concentration is very low ( say, about 10^-6 M) the contribution of water to [H+] becomes appreciable and cannot be neglected .

H3O+ + OH -

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