Question

When the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is

Answer 1

PH = -log( [H3O+] )
so the pH is in powers of 10
1* 10^ 5 / 1*10^3 = 1* 10^2 = 100
 so the answer is:
D. 100 times the original content

Answer 2

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.