Question

You have 3 beakers:

Beaker 1 contains 50.0 ml of 0.123 M HC2H3O2

Beaker 2 contains 25.0 ml of 0.456 M NaC2H3O2

Beaker 3 contains 5.0 ml of 0.789 M HCl

A.) What is the pH of the solution in beaker 1?

B.) The contents of beaker 1 are poured into Beaker 2 and mixed. What is the pH of the resulting solution?

C.) The contents of beaker 3 are now poured into the resulting solution. What is the new pH?

D.) Could this final solution (part c) be an effective buffer?

Answer 1

A.

               CH3COOH           <------->           H⁺           +          CH₃COO^-
IC:              0.123                                         0                                  0
C:                  - x                                         + x                             + x
EC:          0.123 – x                                       x                                 x

Ka = [H⁺] [CH₃COO^-] / [CH₃COOH]

Ka of CH3COOH = 1.8 x 10^-5

So,
1.8 x 10^-5 = [H⁺] [CH₃COO^-] / [CH₃COOH]
1.8 x 10^-5 = (x) (x) / (0.123 –x)
1.8 x 10^-5 = x² / (0.123)           [since Ka is very small, x term in denominator can be discarded)
x² = 2.2 x 10^-6
x = 1.48 x 10^-3

So, [H⁺] = x = 1.48 x 10^-3

pH = - log[H⁺]
      = - log (1.48 x 10^-3)
      = 2.83

(B)

50.0 mL of 0.123 M CH₃COOH
Moles of CH₃COOH = 0.123 M x 0.050 L = 0.00615 moles

25.0 mL of 0.456 M CH₃COONa
Moles of CH₃COONa = 0.456 M x 0.025 L = 0.0114 moles

Total volume = 50 mL + 25 mL = 75 mL = 0.075 L

[CH₃COOH] = 0.00615 moles / 0.075 L = 0.082 M
[CH₃COONa] = 0.0114 moles / 0.075 L = 0.152 M

Using Henderson-Hessalbalach equation

pH = pKa + log {[CH₃COONa] / [CH₃COOH] }

Ka of CH3COOH = 1.8 x 10^-5

pKa = - log Ka
       = - log (1.8 x 10^-5)
       = 4.74

pH = pKa + log {[CH₃COONa] / [CH₃COOH] }

pH = 4.74 + log {0.152 / 0.082 }

pH = 4.74 + log (1.85)

pH = 4.74 + 0.27

pH = 5.01

(c)

[CH₃COOH] = 0.082 M
[CH₃COONa] = 0.152 M

Volume = 75 mL

So,
Moles of CH₃COOH = 0.082 M x 0.075 L = 0.00615 mole
Moles of CH₃COONa = 0.152 M x 0.075 L = 0.0114 moles

5.0 mL of 0.789 M HCl
Moles of HCl = 0.789 M x 0.005 L = 0.003945 moles

So, 0.003945 moles of HCl will react with 0.003945 moles of CH₃COONa to produce 0.003945 moles of CH₃COOH. So, moles of CH₃COONa will decrease and moles of CH₃COOH will increase.

Moles after addition of HCl
Moles of CH₃COOH = 0.00615 mole + 0.003945 moles = 0.010095 moles
Moles of CH₃COONa = 0.0114 moles - 0.003945 moles = 0.007455 moles

Total volume = 75 mL + 5 mL = 80 mL = 0.080 L

So,
[CH₃COOH] = 0.010095 moles / 0.080 L = 0.1261875 M
[CH₃COONa] = 0.007455 moles / 0.080 L = 0.0931875 M

Using Henderson-Hessalbalach equation

pH = pKa + log {[CH₃COONa] / [CH₃COOH] }

Ka of CH3COOH = 1.8 x 10^-5

pKa = - log Ka
       = - log (1.8 x 10^-5)
       = 4.74

pH = pKa + log {[CH₃COONa] / [CH₃COOH] }

pH = 4.74 + log {0.0931875 / 0.1261875 }

pH = 4.74 + log (0.738484398)

pH = 4.74 - 0.13

pH = 4.61

(d)

Yes, this final solution is still buffer. Since pH < pKa.

Hence, it has not reached the half equivalent point.