10 {}^( - 16) M \: HCL \text{ \: is \: divided \: to \: 100 \: times \: its \: pH \: will \: be} Please Help me with this 25 Points_​

Question

`10 {}^( - 16) M \: HCL \text{ \: is \: divided \: to \: 100 \: times \: its \: pH \: will \: be}`

Please Help me with this
25 Points_​

Answer 1

`\text{Given, Concentration of HCl}= 10 {}^( - 6) M`

`\text{After Dilution conc. of HCl =} \frac{10 {}^( - 6) }{100} = {10}^( - 8)M`

`\therefore \text{pH = - log[H}^( + ) ]`

`\therefore \text{ pH }= - \log [{} {10}^( - 8)] = 8`

`\text{But} \: \text{This} \: \text{Is} \: \text{Not True Because} \\ \text{ An Acidic Solution Cannot have} \text{pH Greater than 7.} \\ \text{ In this case} [ {H}^( + ) ] \text{ Ions Of Water Molecule} \\ \: \text {Cannot Be Neglected.}`

Thus,

`{[H}^( + ) ] = {[H}^( + ) ] _(HCl) +{ [H + }^( ) ] _(H_2 O)`

As we know, ionic product

`\text{Now from, pH} = - \text{log}[H {}^( + ) ] \\ = \text{ - log}(11 * {10}^( - 8) ) \\ = - \text{ log}11 - \text{ log} {10}^( - 8) \\ = 6.957`

Hope This Helps!!