PART 1: For 500.0 mL of a buffer solution that is 0.155 M in HC2H3O2 and...

PART 1: For 500.0 mL of a buffer solution that is 0.155 M in HC2H3O2 and 0.145 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

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PART 2: For 500.0 mL of a buffer solution that is 0.150 M in CH3CH2NH2 and 0.135 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

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Expert Solution

Part-A

Ka = 1.76x 10^-5
pKa = 4.75
pH = 4.75 + log (0.145 / 0.155) =4.72 ( initial pH of the buffer )
moles acetic acid = 0.155 M x 0.5 L= 0.0775
moles acetate = 0.145 M x 0.5 L = 0.0725
the effect of the added 0.020 mol H+ would be to increase the moles of acetic acid by 0.020 and decreases the moles of acetate by 0.020 by the reaction
CH3COOH + H+
moles CH3COOH = 0.0775 + 0.020 = 0.0975
moles CH3COO- = 0.0725 - 0.020= 0.0525
concentration CH3COOH = 0.0975 / 0.5 L= 0.195 M
concentration CH3COO- = 0.0525 / 0.5 = 0.105
pH = 4.72 + log 0.105 / 0.195 = 4.45( final pH)

Part - B

Kb of ethylammine = 4.3 x 10^-4
pKb = 3.37
pOH = 3.37 + log 0.135 / 0.15 = 3.325
pH = 14 - 3.325 =10.675 ( initial pH)
moles CH3CH2NH3+ = 0.150 x 0.5 L= 0.075
moles CH3CH2NH2 = 0.135 x 0.5 L=0.0675
CH3CH2NH2 + H+ = CH3CH2NH3+
moles CH3CH2NH3+ = 0.075 + 0.02 = 0.095
moles CH3CH2NH2 = 0.0675 - 0.02= 0.0475
concentration CH3CH2NH3+ = 0.095 / 0.5 = 0.19 M
concentration CH3CH2NH2 = 0.0475 / 0.5= 0.095 M
pOH = 3.37 + log 0.19 / 0.095= 3.07
pH = 10.93

queen_honey_blossom answered 1 year ago

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