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12 Experiment 25 pH Measurements-Buffers and Their Properties Formic acid. a pH of 3.55 from HFor, has a K, value equal to about l .8 × 10 4. A student is asked to prepare a buffer having a solution of formic acid and a solution of sodium formate having the same molarity milliliters of the NaFor solution should she add to 20. ml. of the HFor solution to make the 4, o buffer? (See discussion of buffers.) 097 How many ml of 0.10 M NaOH should the student add to 20 mL 0.10 M HFor if she wished to prepare a buffer with a pH of 3.55, the same as in Problem 4? mL.

Explanation / Answer

4. According to Henderson-Hasselbulch equation: pH = pKa + Log([NaFor]*VNaFor/[HFor]*VHFor), where V is the corresponding volume.

i.e. 3.55 = -Log(1.8*10-4) + Log(VNaFor/20)

i.e. 3.55 = 3.745 + Log(VNaFor/20)

i.e. Log(20/VNaFor) = 3.745-3.55 = 0.195

i.e. 20/VNaFor = 100.195 = 1.566

i.e. VNaFor = 20/1.566 = 12.77 mL

5. According to Henderson-Hasselbulch equation: pH = pKa + Log([NaOH]*VNaOH/[HFor]*VHFor - [NaOH]*VNaOH), where V is the corresponding volume.

i.e. 3.55 = 3.745 + Log(0.1*VNaOH/0.1*20 - 0.1*VNaOH)

i.e. Log(0.1*VNaOH/2 - 0.1*VNaOH) = 0.195

i.e. 0.1*VNaOH/2 - 0.1*VNaOH = 1.566

i.e. 0.1*VNaOH = 3.132 - 0.1566*VNaOH

i.e. 0.2566 VNaOH = 3.132

i.e. VNaOH = 3.132/0.2566 = 12.2 mL

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