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 HClO4 solution. Calculate the pH after the following volumes of acid have been added: 30.0 mL.
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Expert Answer

David Clayton
Answered 2021-12-20 Author has 36 answers

Moles of OHions=Molarity of KOH solution×Volume of sample=0.150M×0.020L=3×103 mole
Moles of H+ions= Molarity of HClO4 solution×Volume of sample=0.125M×0.030L=3.75×103 mole
Reaction:
H+(aq)+OH(aq)H2O(l)
Initial moles of OH and H+ are 3×103 and 0 mole respectively.
Moles of OH and H+ after the reaction are 0 and 0.75×103 respectively.
Molarity of H+ are: 0.75×1030.0500=0.015M
pH=log[H+]=log[0.015]=1.82

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Robert Pina
Answered 2021-12-21 Author has 42 answers
We are given:
- Concentration of KOH=0.150M
- Volume of KOH=20.0mL
- Concentration of HClO4=0.125M
- Volume of HClO4=30mL
Using the given information, we need to determine the pH when the titration is finished. Both KOH and HClO4 are strong base and acid, respectively. Thus, when dissolved in water, all of the OH in the KOH becomes free hydroxide and all of the H+ in the HClO4 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 HClO4 is just a double-replacement reaction forming water and a salt. The reaction is given as:
KOH+HClO4H2O+KClO4
This equation is already balanced and as we can see, 1 mole of KOH reacts with 1 mole of HClO4. Therefore, we can easily determine the excess since KOH and HClO4 react in a 1:1 molar ratio.
Next, lets

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nick1337
Answered 2021-12-27 Author has 575 answers

Volume of HClO4=30mL
Moles of KHClO4=Molarity×volume
=0.125M×30mL=3.75mmol
Left HClO4=(3.753)mmol
Volume of solution =(20+30)mL
Concentration of H+=MolesVolume
=(3.753)mmol50mL=0.015M
pH=log(H+)
=log(0.015)
=1.82

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