You are to prepare 100. mL of an acetate buffer of pH 5.00 using 1.0 M acetic ac

Question

You are to prepare 100. mL of an acetate buffer of pH 5.00 using 1.0 M acetic acid (HC2H302) and solid sodium acetate (NaC2H3O2). To prepare for the investigation, complete the following calculations The pKa for HC2H302 is 4.75. (a) What raiio of INal0l o s roquired? 7 (b) You will want the concentrations relatively dilute, so use an acid concentration of 0.10 M. What volume of 1.0 M HC2H302 would be needed to prepare 100. mL of buffer with an acid concentration (c) How many grams of NaC2H302 3 H2O would be needed to prepare 100. mL of buffer if the acid conceniration is 0.10 M308014 (d) If you add 2.4 mL of 0.50 M NaOH to 25.0 mL of the buffer, what is the new pH? 76 The number is greater than 2% off the expected value

Explanation / Answer

Ans. #a. AH = CH3COOH     ; A- = CH3COO-

Using HH equation-

            pH = pKa + log ([A-] / [AH]

            Or, 5.00 – 4.75 = log ([A-] / [AH]

            Or, ([A-] / [AH] = 100.25 = 1.7782

Therefore, [A-] / [AH] = 1.7782 = 1.8 (Your answer seems to be correct. Please re-check if there any instruction for number of significant figures.)

#c. Given, [AH] = 1.0 M

Calculated [A-] / [AH] = 1.7782

Now,

            [A-] = 1.7782 [AH] = 1.7782 x 1.0 M = 1.7782 M

# 1 mol CH3COO- ( = A-) is produced by 1 mol CH3COONa. So, required moles of sodium acetate is the same as the required moles of acetate ion.

So,

Required mass of CH3COONa = (Molarity x Vol. in liters) x MW

                                                = (1.7782 M x 0.100 L) x (82.034388 g/ mol)

                                                = 14.5874 g

#d. Total volume of buffer after NaOH addition = 2.0 mL + 25.0 mL = 27.0 mL

# [NaOH] in reaction mixture, C2 = (0.50 M x 2.4 mL) / 27.0 mL = 0.0444 M

# Initial [AH] in reaction mixture, C2 = (1.0 M x 25.0 mL) / 27.0 mL = 0.9259 M

# Initial [A-] in reaction mixture, C2 = (1.7782 M x 25.0 mL) / 27.0 mL = 1.6465 M

# Balanced reaction:             AH + OH- ----> A- + H2O

1 mol NaOH neutralizes 1 mol AH to form 1 mol A?

# [AH] neutralized = [NaOH] in Rxn mixture = [A-] formed = 0.0444 M

Equilibrium [AH] = Initial [AH] – [AH] neutralized = 0.9259 M – 0.0444M = 0.8815 M

Equilibrium [A-] = Initial [A-] – [A-] formed = 1.6465 M + 0.0444M = 1.6909 M

Now,

            pH of the final buffer = 4.75 + log (1.6909 / 0.8815) = 4.75 + 0.27 = 5.02

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