Question

Question #46084

Answer 1

`["H"_3"O"^(+)] = 5.6 * 10^(-8)"M"`

Explanation:

As you know, a solution's pH is a measure of the concentration of hydronium cations, `"H"_3"O"^(+)`, found in that solution. More specifically, a solution's pH is equal to

`color(blue)(|bar(ul(color(white)(a/a)"pH" = - log( ["H"_3"O"^(+)])color(white)(a/a)|)))`

In order to find the concentration of the hydronium cations when given the pH of the solution, use the fact that

`color(purple)(|bar(ul(color(white)(a/a)color(black)(10^log(a) = a)color(white)(a/a)|)))`

Rearrange the equation that defines the pH to get

`-"pH" = log(["H"_3"O"^(+)])`

This will be equivalent to

`10^(-"pH") = 10^log(["H"_3"O"^(+)])`

Finally, rearrange to find

`color(blue)(|bar(ul(color(white)(a/a)["H"_3"O"^(+)] = 10^(-"pH")color(white)(a/a)|)))`

Plug in the value given to you for the pH of the solution to get

`["H"_3"O"^(+)] = 10^(-7.25) = color(green)(|bar(ul(color(white)(a/a)5.6 * 10^(-8)"M"color(white)(a/a)|)))`

It's worth noting that a neutral aqueous solution at room temperature has equal concentrations of hydronium and hydroxide ions

`["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M"`

As you can see, decreasing the concentration of hydronium ions results in an increase in pH. As a result, the solution goes from being neutral to being slightly basic.