Question

In: Chemistry

Part B) For 490.0 mL of a buffer solution that is 0.175 M in HC2H3O2 and...

Part B) For 490.0 mL of a buffer solution that is 0.175 M in HC2H3O2 and 0.155 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.020 mol of HCl. Part C) For 490.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.165 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Expert Solution

Part B) For 490.0 mL of a buffer solution that is 0.175 M in HC2H3O2 and 0.155 M in NaC2H3O2, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

Here HC2H3O2 = acetic acid

Ka = 1.75 × 10 ⁻⁵
pKa = 4.75

initial pH of this buffer:

pH = pKa+ log salt / acid

pH = 4.75+ log 0.155 / 0.175

=4.75-0.053

=4.697

moles acid = 0.175 M x 0.490 L=0.08575
moles salt = 0.155 M x 0.490 L =0.07595

Moles of HCl= 0.020 mol of HCl

0.07595-0.020= 0.05595 mol of the base of the buffer left in solution

0.08575+0.020= 0.10595 mol of the acid of the buffer now in solution
now final pH:

pH = 4.75 (pka of weak acid) + log (0.05595/0.10595)

=4.75-0.277

= 4.473

Part C) For 490.0 mL of a buffer solution that is 0.185 M in CH3CH2NH2 and 0.165 M in CH3CH2NH3Cl, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

initial pH:

Kb of Ethyl Amine = 5.6*10^4

pKb = 3.25

pOH = pKb + log(BA-/Base_

pOH =3.25+ log(0.165/0.185)

pOH = 3.25-0.05

pOH = 3.20
pH = 14 -3.20

= 10.80

Here adding HCl will convert the base NH3 into NH4Cl....

CH3CH2NH2 + HCl ------> CH3CH2NH3Cl

number of moles of HCl =0.02 moles

number of moles of CH3CH2NH3Cl in 490 ml = 0.165X 0.490 = 0.08085
number of moles of CH3CH2NH2 in 490 ml = 0.185X0.490= 0.09065

as 1 mole of HCl converts 1 mole of CH3CH2NH2 into 1 mole of CH3CH2NH3Cl

so 0.02 moles of HCl will convert 0.02 moles of CH3CH2NH2 into 0.02 moles of CH3CH2NH3Cl

so now total number of moles of CH3CH2NH2 = 0.09065 -0.02 = 0.07065
number of moles of CH3CH2NH3Cl = 0.08085+0.05 = 0.10085

total volume = 490 ml = 0.490 L

so [salt] = 0.10085/0.490 = 0.206 M
[base] = 0.07065/0.490 = 0.144 M

pOH = pKb + log(BA-/Base_

pOH =3.25+ log(0.144/0.206)
pOH = 3.25-0.156
pOH = 3.104

so pH = 14-3.104 = 10.896

queen_honey_blossom answered 2 years ago

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