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1. A buffer solution consists of 0.25 M acetic acid (CH3COOH) and 0.10 M sodium acetate...

1. A buffer solution consists of 0.25 M acetic acid (CH3COOH) and 0.10 M sodium acetate (CH3COONa). The Ka of acetic acid is 1.8×10−5. a. Calculate the equilibrium pH using the equilibrium mass action expression.

b. Calculate the equilibrium pH using the Henderson-Hasselbalch equation approach.

2. Consider a buffer solution that consists of both benzoic acid (C6H5COOH) and sodium benzoate (C6H5COONa) and has a pH of 4.18. The concentration of benzoic acid is 0.40 M. The pKa of benzoic acid is 4.18. a. What is the concentration of sodium benzoate in the buffer solution?

b. What volume (in mL) of 0.20 M NaOH is required to reach the equivalence point if you start with 10-mL of the benzoic acid/benzoate buffer?

c. How will the volume of 0.20 M NaOH required to reach the equivalence point differ in part (b) if the 10-mL solution of benzoic acid/benzoate buffer has a pH higher than 4.18?

Solutions

Expert Solution

1) a) Equilibrium pH using the equilibrium mass action expression:

           CH3COOH    ----------------> CH3COO- + H+

Initial      0.25 M                                   0             0

At equilibrium         0.25-X                                     X             X

   Then,

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

                      1.8×10−5 = X.X / (0.25-X)

            On solving, X = 0.00211 M

Hence, [H+] = X = 0.00211 M

pH = - log [H+] = - log (0.00211) = 2.67

Therefore,

    pH = 2.67

b) Equilibrium pH using the Henderson-Hasselbalch equation:

We know that

Henderson-Hasselbalch equation is pH = pKa + log [A-]/HA]

                                                    pH = - logKa + log [A-]/HA]

                                                        = - log (1.8×10−5) + log (0.1 / 0.25)

                                                      = 4.34

Therefore,

pH = 4.34

                    

queen_honey_blossom answered 1 year ago
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