# Question 7aa1b

Feb 3, 2016

See explanation.

#### Explanation:

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, ${\text{H"_3"O}}^{+}$.

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 $1$

$\log \left(x\right) < 0 \text{ } \left(\forall\right) \textcolor{w h i t e}{a} 0 < x < 1$

You know that the molarity of the hydronium ions is equal to ${\text{0.2 mol dm}}^{- 3}$, so plugging this into the equation will get you

$\text{pH} = - \log \left(0.2\right) = - \left(- 0.699\right) = 0.7$

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

${10}^{{\log}_{10} \left(x\right)} = x$

$\text{pH} = 0.7$

you can say that

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

$\log \left(\left[{\text{H"_3"O}}^{+}\right]\right) = - 0.7$

Since you can say that if $a = b$, then ${10}^{a} = {10}^{b}$, you will have

${10}^{\log} \left(\left[{\text{H"_3"O}}^{+}\right]\right) = {10}^{- 0.7}$

This will get you

$\left[{\text{H"_3"O}}^{+}\right] = {10}^{- 0.7} = 0.1995 = 0.2$