Need to prepare buffer with Henderson-Hasselbalch Equation?

Question: Use the H-H equation to calculate the volume of 0.1 M acetic acid and 0.1 M Na acetate required to prepare 50 ml 0.1 M acetate buffer pH 4.0. HH: pH = pk + log[base/acid] Acetic acid is an acid but acetate is not a base. Acetate is a weak acid but I put it as a base bc how am I suppose to use HH to get a ratio? Here is what I tried: pH = pka + log[0.1 M acetate/0.1 M acetic acid] 4.0 - 4.76 = log[0.1 M Na acetate/0.1 M acetic acid] 0.76= log[0.1M Na acetate/0.1 M acetic acid] 5.75 = [0.1 M Na acetate/0.1 M acetic acid] 5.75 = [0.1 NaOH]/[0.1 acetic acid] [acetic acid] = NaOH/5.74 Vol of acetic acid = (50 ml) [NaOH]/5.74 = 8.71 ml Vol of acetate = 50 ml - 8.71 = 41.3 ml However, these are wrong answers. The correct answers should be: 42.6 ml 0.1 M acetic acid + 7.4 ml 0.1 M Na acetate to give pH 4.0. Tell me what i did wrong and how to derive at the correction solutions above. Thanks.

Answer: 4.00 - 4.76 = - 0.76 10^-0.76 = 0.173 = [acetate ] / [acetic acid] Since both acetate and acetic acid have the same concentration this means that we take 0.173 L of acetate per L of acetic acid . Total volume = 1.173 L = 1173 mL 173 : 1173 = x : 50 x = mL of acetate = 7.4 mL of acetic acid = 50 - 7.4 = 42.6 mL I hope this help you

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