Question #7aa1b

I'm not really sure what you're asking here, so I'll just do a quick breakdown of how the pH of a solution is calculated.

As you know, the pH of the solution is simply a measure of the concentration of hydronium ions, #"H"_3"O"^(+)#.

More specifically, the pH of a solution is calculated by taking the negative log base 10 from the concentration of hydronium ions.

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

You need to take the negative log of the concentration because you're dealing with concentrations that are are, in most cases, smaller than #1#

#log(x) < 0" " (AA) color(white)(a) 0 < x < 1#

You know that the molarity of the hydronium ions is equal to #"0.2 mol dm"^(-3)#, so plugging this into the equation will get you

#"pH" = - log(0.2) = - (-0.699) = 0.7#

You can find the concentration of hydronium ions by using the fact that

#10^(log_10(x)) = x#

So if you start with

#"pH" = 0.7#

you can say that

#overbrace(-log(["H"_3"O"^(+)]))^(color(purple)(="pH")) = 0.7#

#log(["H"_3"O"^(+)]) = - 0.7#

Since you can say that if #a = b#, then #10^a = 10^b#, you will have

#10^log(["H"_3"O"^(+)]) = 10^(-0.7)#

This will get you

#["H"_3"O"^(+)] = 10^(-0.7) = 0.1995 = 0.2#