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Please fill in the following table. Thanks for all the help! Give the pKa and pK

ID: 487633 • Letter: P

Question

Please fill in the following table. Thanks for all the help!

Give the pKa and pKb of weak acids and bases.
For each solution, calculate the pH of a 0.1000 M solution, and calculate the volume of 0.1000 M NaOH or HCl that would be required to titrate a 10.00 mL sample to the equivalence point.
To fill-in the table below, answer the subsequent questions.

pKb(NH3) =  

pKa(CH3COOH) =  

pKa(NH4Cl) =  

pKb(CH3COONa) =  

pH(0.1000 M HCl) =  

pH(0.1000 M KOH) =  

pH(0.1000 M NH3) =  

pH(0.1000 M CH3COOH) =  

ptspH(0.1000 M NH4Cl) =

pH(0.1000 M CH3COONa) =  

Volume of 0.1000 M NaOH required to titrate 10.00 mL of 0.1000 M HCl =  

Volume of 0.1000 M HCl required to titrate 10.00 mL of 0.1000 M KOH =  

Volume of 0.1000 M HCl required to titrate 10.00 mL of 0.1000 M NH3 =  

Volume of 0.1000 M NaOH required to titrate 10.00 mL of 0.1000 M CH3COOH =  

Volume of 0.1000 M NaOH required to titrate 10.00 mL of 0.1000 M NH4Cl =  

Volume of 0.1000 M HCl required to titrate 10.00 mL of 0.1000 M CH3COONa =  

Compound acid or base initial pH type pKa pKb strong acid HCI KOH strong base NH3 weak base CH3COOH weak acid NH4Cl weak acid CH3COON a weak base volume titrant (mL)

Explanation / Answer

a)

pKa for HCl --> N/A since strong acid

pH initially 0.1 M

[H+] =[Hcl] = 0.1

pH = -log(0.1) =1

Volume required = Macid*Mbase / Mbase = 0.1*10 / 0.1 = 10 mL of base required

b.

pKb for KOH--> N/A since strong base

pH initially 0.1 M

[OH-] =[KOH] = 0.1

pOH = -log(OH) = -log(0.1) = 1

pH = 14-1 = 13

Volume required = Mbase*Mbase / Macid = 0.1*10 / 0.1 = 10 mL of base required

c)

NH3

pKb = 4.75

initially

NH3 + H2O <-> NH4+ + OH-

Kb = [NH4+][OH-]/[NH3]

10^-4.75 = x*x / (0.1-x)

x = OH = 0.00132

pOH = -log(0.00132) = 2.879

pH = 14-2.879 = 11.121

volume required:

Volume required = Mbase*Mbase / Macid = 0.1*10 / 0.1 = 10 mL of base required

c)

for acetic acid:

pKa = 4.75

CH3COOH <-> CH3COO- + H+

Ka = [CH3COO-][H+]/[CH3COOH ]

1.8*10^-5 = x*x/(0.1-x)

x = H+= 0.00132

pH= -log(0.00132) = 2.879

volume required:

Volume required = Macid*Vacid/ Mbase= 0.1*10 / 0.1 = 10 mL of base required

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