Calculate the pH of each of the following mixtures.   500 ml 0.35 M KOH + 500 ml 0.25 M HNO3

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Calculate the pH of each of the following mixtures.

 

500 ml 0.35 M KOH + 500 ml 0.25 M HNO3

500 ml 0.35 M CH3CH2NH2 + 500 ml 0.25 M HNO3

500 ml 0.35 M HCl + 500 ml 0.25 M CH3CH2NH2

500 ml 0.35 M HF + 500 ml 0.25 M KOH

Expert Solution
Step 1 Let us consider the following reaction for 1st mixture

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The mixture will go through a neutralization reaction

KOH+HNO3KNO3+H2O

 

Step 2 Determination of number of moles of KOH and HNO3

Molarity of KOH=0.35 M

Given Volume of KOH= 500ml = 0.5 L

So, Number of moles of KOH =  Molarity x Volume = 0.35 x 0.5 = 0.175 mol

Molarity of HNO3 = 0.25 M

Given Volume = 500ml = 0.5 L

So, Number of moles of HNO3 = Molarity x Volume = 0.25 x 0.5= 0.125 mol

We can see that, Number of moles of KOH > Number of moles of HNO3 

Therefore, HNO3 is the limiting reagent and there will be KOH remaining.

The reaction will continue until all HNO3 is consumed

Number of moles of KOH remaining = 0.175-0.125 = 0.050 moles

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