Calculate the pH of 0.100 L of the buffer 0.140 M CH3COONa/0.120 M CH3COOH befor

Question

Calculate the pH of 0.100 L of the buffer 0.140 M CH3COONa/0.120 M CH3COOH before and after the addition of the following species. (Assume there is no change in volume.)

(a) pH of the starting buffer:

(b) pH after addition of 0.0030 mol HCl

(c) pH after addition of 0.0020 mol NaOH (added to a fresh solution of the starting buffer):

2. Solid NaI is slowly added to a solution that is 0.0087 M Cu+ and 0.0083 M Ag+.

Which compound will begin to precipitate first?

NaI

CuI

AgI

Calculate [Ag+] when CuI just begins to precipitate.

__×10__ M Enter your answer in scientific notation.

What percent of Ag+ remains in solution at this point?

%

NaI

CuI

AgI

Explanation / Answer

Here we use the "Henderson-Hasselbalch equation" to solve for the pH of the solution.

So, according to the Henderson-Hasselbalch equation,

pHsol = pKa + log ( [COONa] [CH3COONa] ) ........1)

The acid dissociation constant [pKa] for CH3COONa = 1*10^4 [ for weak acid generally low]

pKa = log(Ka) = log( 1 * 104) = 4

put this value in equation 1),

pHsol = 4 + log (0.12/0.14) = 4 + log (0.857) = 4 - 0.067 = 3.933 pH of the starting buffer

b) HCL + CH3COOH --> CH3COCl + H2O

Need to calculate how many moles of CH3COOH and of CH3COONa you initially had in the buffer.

C=nVn=CV

nCH3COOH = 0.12 M0.100 L = 0.012 moles

nCH3COONa = 0.14 M0.100 L = 0.014 moles

Calculate the new molarities of the species present in the buffer after addition of 0.0030 mol HCl

[CH3COOH] = (0.012 - 0.003) / 0.1 = 0.09 M

[CH3COONa] = (0.014 + 0.003) / 0.1 = 0.17 M

Use equation (1) to determine the new pH

pHsol 2 = 4 + log (0.09/0.15) = 3.778

c) Calculate the new molarities of the species present in the buffer after addition of 0.0020 mol NaOH

  [CH3COOH] = (0.012 - 0.002) / 0.1 = 0.1 M

[CH3COONa] = (0.014 + 0.002) / 0.1 = 0.16 M

pHsol 3 = 4 + log (0.1/0.16) = 3.795

2. AgI precipitates first because it has lower Ksp and lower solubility:

Ksp =[ Ag+][I-] = 1.5x10^-16
[0.010][I-] = 1.5x10^-16
[I-] = 1.5x10^-14    when AgI begins to precipitate.

Ksp = [Cu+][i-] = 5.3x10^-12 for CuI
[0.010][I-] = 5.3x10^-12
[I-] = 530x 10^-12 when CuI begins to precipitate.

Now,

[Ag+] [5.32x10^-10] = 1.5x10^-16

  [Ag+] = 0.28x10^-6

% Ag+ in solution = (100) [Ag+] / (0.010)

= (100x0.28x10^-6) /0.010

= 2800 x 10^-6

=0.0028 %

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