# How to determine the pH of the following mixture and the amount of salt formed?

## 250 mL HBr 0,1 M + 250 mL KOH 0,2 M

Feb 10, 2018

$\text{pH} = 12.7$

#### Explanation:

Hydrobromic acid and potassium hydroxide react in a $1 : 1$ mole ratio to produce aqueous potassium bromide and water.

${\text{HBr"_ ((aq)) + "KOH"_ ((aq)) -> "KBr"_ ((aq)) + "H"_ 2"O}}_{\left(l\right)}$

This means that if you mix equal numbers of moles of each reactant, you will have a complete neutralization, which implies that both the acid and the base will be completely consumed.

However, in your case, you do not have a complete neutralization because you're mixing twice as many moles of potassium hydroxide than of hydrobromic acid.

overbrace(250 color(red)(cancel(color(black)("mL"))) * "0.2 moles KOH"/(10^3color(red)(cancel(color(black)("mL")))))^(color(blue)("moles KOH")) = 2 * overbrace(250 color(red)(cancel(color(black)("mL"))) * "0.1 moles HBr"/(10^3color(red)(cancel(color(black)("mL")))))^(color(blue)("moles HBr"))

More specifically, you're mixing $0.050$ moles of hydrobromic acid and $0.025$ moles of potassium hydroxide.

In this case, hydrobromic acid will act as the limiting reagent, i.e. it will be completely consumed before all the moles of potassium bromide will get the chance to react.

After the reaction is complete, the resulting solution will contain $0$ moles of hydrobromic acid and

$\text{0.050 moles " - " 0.025 moles" = "0.025 moles KOH}$

This is the case because when $0.025$ moles of hydrobromic acid are consumed, $0.025$ moles of potassium hydroxide are consumed as well.

Moreover, the resulting solution will also contain $0.025$ moles of aqueous potassium bromide because this product is produced in $1 : 1$ mole ratios with the two reactants.

The total volume of the resulting solution will be

$\text{250 mL + 250 mL = 500 mL}$

This means that the concentration of hydroxide anions, whch are delivered to the solution by the potassium hydroxide in a $1 : 1$ mole ratio, will be equal to

["OH"^(-)] = "0.025 moles"/(500 * 10^(-3) quad "L") = "0.05 M"

As you know, an aqueous solution at ${25}^{\circ} \text{C}$ has

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{pH + pOH = 14}}}}$

You also know that

color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)])))

so you can say that the $\text{pH}$ of the resulting solution will be

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

Plug in your value to find

$\text{pH} = 14 + \log \left(0.05\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{12.7}}}$

The answer is rounded to one decimal place because you have one significant figure for the concentrations of the two solutions.

Finally, to find the mass of aqueous potassium bromide produced by the reaction, use the compound's molar mass

$0.025 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles KBr"))) * "119.002 g"/(1color(red)(cancel(color(black)("mole KBr")))) = color(darkgreen)(ul(color(black)("3 g}}}}$

The answer is rounded to one significant figure.