Question

Question #36e0b

Answer 1

`"pH" = 11`

Explanation:

The thing to remember about an aqueous solution at room temperature is that

`color(blue)(ul(color(black)(["OH"^(-)] * ["H"_ 3"O"^(+)] = 10^(-14))))`

Rearrange this equation to isolate the concentration of hydronium cations on one side

`["H"_ 3"O"^(+)] = (10^(-14))/(["OH"^(-)])`

Now, you should be aware that

`color(blue)(ul(color(black)("pH" = - log( ["H"_3"O"^(+)]))))`

Plug the expression you have for the concentration of the hydronium cations in this equation to get

`"pH" = - log( (10^(-14))/(["OH"^(-)]))`

In your case, you will have

`"pH" = -log( (10^(-14))/(10^(-3)))`

`"pH" = - log(10^(-11))`

`"pH" = -(-11) log(10)`

`color(darkgreen)(ul(color(black)("pH" = 11)))`