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  • Solve this problem The PH of boiling water (373k) is ............(Kwat 373K=10-12)

The PH of boiling water (373k) is ............(Kw at 373K=10-12)

  • Option 1)

    12

  • Option 2)

    8

  • Option 3)

    6

  • Option 4)

    2

 

Answers (1)

best_answer

As we learned in concept

p(Kw) -

K_{w}=[H_{2}O^{+}]\:[H\bar{O}]=10^{-14}

Although  K_{w}  may change with temperature the variation in pH with temperature are so small that we often ignore it.

- wherein

p(K_{w})=-logK_{w}

               =-log{[H_{3}O^{+}]\:[\bar{O}H]}

               =-log10^{-14}

p(K_{w})=p(H)+p(OH)

                 =14

 

 K_{w}=10^{-12}

=>\left [ H^{+} \right ]\left [ OH^{-} \right ]=10^{-12}

=>\left [ H^{+} \right ]^{2}=10^{-12}

=>\left [ H^{+} \right ]=10^{-6}

pH=-log\left [ H^{+} \right ]=-log(10^{-6})=6


Option 1)

12

Option is incorrect

Option 2)

8

Option is incorrect

Option 3)

6

Option is correct

Option 4)

2

Option is incorrect

Posted by

prateek

View full answer

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