# A 20.0-mL sample of 0.150 M KOH is titrated with

Mary Buchanan 2021-12-19 Answered
A 20.0-mL sample of 0.150 M KOH is titrated with 0.125 M $HCl{O}_{4}$ solution. Calculate the pH after the following volumes of acid have been added: 30.0 mL.
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

David Clayton

Moles of $O{H}^{-}$ions=$\text{Molarity of KOH solution}×\text{Volume of sample}=0.150M×0.020L=3×{10}^{-3}$ mole
Moles of ${H}^{+}$ions=  mole
Reaction:
${H}^{+}\left(aq\right)+O{H}^{-}\left(aq\right)⇒{H}_{2}O\left(l\right)$
Initial moles of $O{H}^{-}$ and ${H}^{+}$ are $3×{10}^{-3}$ and 0 mole respectively.
Moles of $O{H}^{-}$ and ${H}^{+}$ after the reaction are 0 and $0.75×{10}^{-3}$ respectively.
Molarity of ${H}^{+}$ are: $-\frac{0.75×{10}^{-3}}{0.0500}=0.015M$
$pH=-\mathrm{log}\left[{H}^{+}\right]=-\mathrm{log}\left[0.015\right]=1.82$

We have step-by-step solutions for your answer!

Robert Pina
We are given:
- Concentration of $KOH=0.150M$
- Volume of $KOH=20.0mL$
- Concentration of $HCl{O}_{4}=0.125M$
- Volume of $HCl{O}_{4}=30mL$
Using the given information, we need to determine the pH when the titration is finished. Both KOH and $HCl{O}_{4}$ are strong base and acid, respectively. Thus, when dissolved in water, all of the $O{H}^{-}$ in the KOH becomes free hydroxide and all of the ${H}^{+}$ in the $HCl{O}_{4}$ becomes free hydrogen ions. The pH will then be determined whichever of the free ions is in excess after forming water.
The neutralization reaction between KOH and $HCl{O}_{4}$ is just a double-replacement reaction forming water and a salt. The reaction is given as:
$KOH+HCl{O}_{4}⇒{H}_{2}O+KCl{O}_{4}$
This equation is already balanced and as we can see, 1 mole of KOH reacts with 1 mole of $HCl{O}_{4}$. Therefore, we can easily determine the excess since KOH and $HCl{O}_{4}$ react in a 1:1 molar ratio.
Next, lets

We have step-by-step solutions for your answer!

nick1337

Volume of $HCl{O}_{4}=30mL$
Moles of $KHCl{O}_{4}=Molarity×volume$
$=0.125M×30mL=3.75mmol$
Left $HCl{O}_{4}=\left(3.75-3\right)mmol$
Volume of solution $=\left(20+30\right)mL$
Concentration of ${H}^{+}=\frac{Moles}{Volume}$
$=\frac{\left(3.75-3\right)mmol}{50mL}=0.015M$
$pH=-\mathrm{log}\left({H}^{+}\right)$
$=-\mathrm{log}\left(0.015\right)$
$=1.82$

We have step-by-step solutions for your answer!