Question

What is the [h+] if the ph of a solution is 11.13

pls help

Answer 1

Step-by-step explanation:

`[{H}^( + )] = {10}^( - pH)`

`= {10}^( - 11.13)`

`= {10}^( (- 11 - 0.13))`

`= {10}^( - 0.13) * {10}^( - 11)`

`= 0.741 * {10}^( - 11) (mol)/(litre)`

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