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.
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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
Reference link will be
https://brainly.in/question/13360093
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