#### How many litres of water must be added to 1 litre of aqueous solution of HCl with a pH of 1 to create an  aqueous solution with pH of 2 ?Option: 1 9.0 LOption: 2 0.1 LOption: 3 0.9 LOption: 4 2.0 L

As we discussed in the concept

pH of Solutions: Strong Acids -

pH is also referred to as potential or power of hydrogen. Mathematically, it can be represented as follows:

If solution is neutral, then:
Kw = [H3O+][OH-]
From the ionic product of water, we know:
Kw = 10-14
[H3O+] = [OH-] = x (since solution is neutral)
Thus, 10-14 = Kw = x2
x = 10-7
Now, [H3O+] = 10-7
Thus, pH = - log10(H3O+) = - log10(10-7) = 7

For Acidic solutions:                                                                                For Basic solutions:
For acidic solutions, we must have [H3O+] > [OH-]                                       For basic solutions, we must have [H3O+] < [OH-
Thus, [H3O+] > 10-7                                                                                   Thus, [H3O+] < 10-7
Thus, [H3O+] for acids can be 10-6, 10-5, 10-4, etc.                                      Thus, [H3O+] for basics can be 10-8, 10-9, 10-10, etc.
Thus, pH of acids can be 6, 5, 4, etc.                                                          Thus, pH of basics can be 8, 9, 10, 11, etc.
Hence, pH of acidic solutions is less than 7                                                  Hence, pH of basic solutions is greater than 7

pH depends upon temperature
We know from ionic product of water that at 630C, the value of Kw = 10-13.
For neutral solution we know:

Hence, pH depends upon temperature

pH of Strong Acids
Strong acids are those acids which dissociate completely in solutions. For example:

• 2 x 10-3 M HNO3
Since HNO3 is a strong acid, thus it will dissociate completely into H+ and OH- ions as follows:

Thus, pH of HNO3 is 2.7

• 10-4 M H2SO4
Since H2SO4 is a strong acid, thus it will dissociate completely into H+ and OH- ions as follows:

Thus, pH of H2SO4 is 3.7

NOTE: If molarity(N) of solution is not given but normality(N) is given, then molarity can be calculated using the following formula:

N = M x n
where, n is the number of moles

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Initial PH

V=10L