# Question d523c

May 10, 2017

$\text{pH} = 10.91$

#### Explanation:

For starters, you should know that

"pOH" = - log(["OH"^(-)])

and that, at room temperature, an aqueous solution has

$\text{pH + pOH} = 14$

This means that you have

$\text{pH" = 14 - "pOH}$

which is equivalent to

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

"pH" = 14 + log(["OH"^(-)])

Now, the initial solution is being diluted by a factor of

"DF" = (750 color(red)(cancel(color(black)("mL"))))/(4.8color(red)(cancel(color(black)("mL")))) = color(blue)(156.25) -> the dilution factor

This means that its concentration must decrease by a factor of $\textcolor{b l u e}{156.25}$. Remember, this is the case because a dilution does not change the number of moles of solute present in the sample.

This implies that you are decreasing the concentration of the solution by increasing the volume of the solution.

The new concentration of the solution will be

${c}_{\text{new" = "0.127 M}} / \textcolor{b l u e}{156.25}$

${c}_{\text{new" = "0.0008128 M}}$

Potassium hydroxide is a strong base, which means that it ionizes completely in aqueous solution to produce potassium cations and hydroxide anions

${\text{KOH"_ ((aq)) -> "K"_ ((aq))^(+) + "OH}}^{-}$

Since every $1$ mole of potassium hydroxide dissolved in water produces $1$ mole of hydroxide anions, you can say that

["OH"^(-)] = ["KOH"] = "0.0008128 M"

Therefore, the $\text{pH}$ of the solution will be equal to

"pH" = 14 + log(0.0008128) = color(darkgreen)(ul(color(black)("10.91)))#

The answer is rounded to two decimal places, the number of sig figs you have for the two volumes.