Question #07916
1 Answer
Here's why that is the case.
Explanation:
That is simply how pH is defined.
You see, pH is actually a measure of concentration. More specifically, a solution's pH will tell you that solution's concentration of hydronium ions,
The pH is calculated by taking the negative log base 10 of the concentration of hydronium ions
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
Since you're dealing with the negative log, solutions that have a relatively high concentration of hydronium ions, i.e. acidic solutions, will have a low pH.
Likewise, solutions that contain a relatively low concentration of hydronium ions, i.e. basic solutions, will have a high pH.
These concentrations are relative to the concentration of hydroxide anions,
In pure water at room temperature, based on water's self-ionization reaction, you have
#color(purple)(|bar(ul(color(white)(a/a)color(black)(["H"_3"O"^(+)] * ["OH"^(-)] = 10^(-14))color(white)(a/a)|)))" "# and#" "color(purple)(|bar(ul(color(white)(a/a)color(black)(["H"_3"O"^(+)] = ["OH"^(-)] = 10^(-7)"M")color(white)(a/a)|))#
This gives a pH for pure water equal to
#"pH"_"pure water" = - log(10^(-7)) = 7#
Now, your solution has a pH of
Now, set up the equation for pH
#"pH" = - log(["H"_3"O"^(+)])#
Rearrange to get
#log(["H"_3"O"^(+)]) = - "pH"#
To get rid of the log, you can write
#10^(log(["H"_3"O"^(+)])) = 10^(-"pH")#
This will be equivalent to
#color(blue)(|bar(ul(color(white)(a/a)["H"_3"O"^(+)] = 10^(-"pH")color(white)(a/a)|)))#
Now all you have to do to get the answer is plug in the value you have for the pH of the solution
#["H"_3"O"^(+)] = 10^(-9.73) = 1.862 * 10^(-10)#
Rounded to two sig figs, the answer will indeed be
#["H"_3"O"^(+)] = 1.9 * 10^(-10)"M"#
Notice that this solution contains a lower concentration of hydronium ions that pure water, which is why its pH is higher than