Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
\(pH=-\log [H_3O^+]\)
The hydronium ion concentration at pH = 5.
\(5=-\log [H_3O^+]\)
\([H_3O^+]=1\times 10^{-5}M\) ..............(1)
The hydronium ion concentration at pH = 3.
\(3=-\log [H_3O^+]\)
\([H_3O^+]=1\times 10^{-3}M\) ................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
\(\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}\)
\(100\times [H_3O^+]_{original}=[H_3O^+]_{final}\)
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.
