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  • 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

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 L


Option: 2

0.1 L


Option: 3

0.9 L


Option: 4

2.0 L


Answers (1)

best_answer

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:

\mathrm{pH\: =\: -log_{10}[H_{3}O^{+}]}

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:

\\\mathrm{[H_{3}O^{+}]\: =\: [OH^{-}]}\\\\\mathrm{\Rightarrow\: K_{w}\: =\: x^{2}}\\\\\mathrm{\Rightarrow\: x\: =\: \sqrt{10^{-13}}\: =\: 10^{-6.5}}\\\\\mathrm{\Rightarrow\: [H_{3}O^{+}]\: =\: 10^{-6.5}}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(10^{-6.5})\: =\: 6.5}

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:

    \\\mathrm{HNO_{3}\: \rightarrow \: H^{+}\: +\: NO_{3}^{-}}\\\\\mathrm{Thus,\: [H^{+}]\: =\: 2\: x\: 10^{-3}M\: \: \: \: \: \: \: \: \: (given)}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2\: x\: 10^{-3})}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2)\: -\: log_{10}(10^{-3})}\\\\\mathrm{\Rightarrow\: pH\: =\: -0.30\: +\: 3\: =\: 2.7}
    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:

    \\\mathrm{H_{2}SO_{4}\: \rightarrow \: 2H^{+}\: +\: SO_{4}^{2-}}\\\\\mathrm{Thus,\: [H^{+}]\: =\: 2\: x\: 10^{-4}M\: \: \: \: \: \: \: \: \: (given)}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2\: x\: 10^{-4})}\\\\\mathrm{\Rightarrow\: pH\: =\: -log_{10}(2)\: -\: log_{10}(10^{-4})}\\\\\mathrm{\Rightarrow\: pH\: =\: -0.30\: +\: 4\: =\: 3.7}
    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

-

 

 

 Initial PH=2=1log[H^{+}]_{2}

[H^{+}]_{1}\times1=[H^{+}]_{2}\times V

10^{-1}\times 1= 10^{-2}\times V

V=10L

Added water =10-1=9L

 

Posted by

himanshu.meshram

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