Question

the pH of rain water collected in a certain region of a country on a particular day was 5.23 calculate the H+ ion concentration of the rain water​

Answer 1

Answer:

Explanation:Given The pH of rain water collected in a certain region of Maharashtra on particular day was 5.1. Calculate the H⊕ ion concentration of the rain water and its percent dissociation.

So given ph of rain water is 5.1. We need to find the concentration of h + ion.

So we have the formula ph = - log 10 [h3 O+]

Percent dissociation = α x 100  

             From the formula, ph = - log 10 [h 3 O +]

               So log 10 [h2 O+] = - 5.1

                           Now add and subtract 1 we get and also we can write this as

                                                = - 0.1 - 5 + 1 - 1  

                  Grouping this we get (- 5 – 1) + 1 – 0.1

                                                     = - 6 + 0.9  

                                                        = 6.9

                          So antilog of 6.9 is 7.94 x 10^-6 M

        ion concentration of the rain water = 7.94 x 10^-6 M                  

So ph of rain water is due to dissociation of monobasic strong acid.

Therefore percent dissociation = α x 100

                                                    = 7.943 x 10^-6 x 100

                                                    = 7.943 x 10^-6 x 10^2

                                                   = 7.943 x 10^-4

Answer 2

Given: the pH of rain water collected in a certain region of a country on a particular day was 5.23.

To find: We have to find out the concentration of hydrogen ion.

Solution:

We know that pH of a species is directly proportional to concentration of hydrogen ion.

The formula is-

$pH = - log( {H}^{ + } )$

Here given that pH=5.23

So, concentration of hydrogen ion will be-

$5.23 = - log( {H}^{ + } ) \\ log( {H}^{ + } ) = - 5.23$

Taking 10 at both sides we get-

$( {H}^{ + } ) = {10}^{ - 5.23} \\ {H}^{ + } = 5.88$

So, the hydrogen ion concentration of rain water is 5.88.