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various household substances, study the properties of buffer he amount of acetic

ID: 1036853 • Letter: V

Question

various household substances, study the properties of buffer he amount of acetic acid and epare the buffers and test their CE112L-05 Spring 2018 Pre-Lab Questions pH and ? I. During this part #3 of the lab, you will make 2 acetic acid buffer solutions (HC2H3O2/NaC2H3O2) of the same pH (pH 4.0), but with different concentration of the components. Use the Henderson-Hasselbalch equation to perform the following calculations. The K, of acetic acid is 1.8 x 105 Buffer A: Calculate the mass of the solid sodium acetate (NaC2H302) required to mix with 100 ml of 0.1 M acetic acid (HC2H302), to prepare a pH 4 buffer. Record the mass in the table, part 3 of this experiment. a. Buffer B: Calculate the mass of the solid sodium acetate (NaC2H302) required to mix with 100 ml of 1.0 M acetic acid (HC2H3O2), to preparea pH 4 buffer. Record the mass in the table, part 3 of this experiment. b.

Explanation / Answer

no of moles of CH3COOH   = molarity * volume in L

                                      = 0.1*0.1 = 0.01moles

no of moles of CH3COONa = x

PKa = -logKa

        = -log1.8*10^-5

       = 4.75

PH = 4

PH     = Pka + log[CH3COONa]/[CH3COOH]

   4       = 4.75 + logx/0.01

logx/0.1    = 4-4.75

logx/0.01    = - 0.75

   x/0.01     = 10^-0.75

   x/0.01     = 0.1778

   x          = 0.1778*0.01

x             = 0.001778

no of moles of CH3COONa = 0.001778moles

mass of CH3COONa = no of moles * gram molar mass

                                 = 0.001778*82   = 0.1458g

no of moles of CH3COOH   = molarity * volume in L

                                      = 1*0.1 = 0.1moles

no of moles of CH3COONa = x

PKa = -logKa

        = -log1.8*10^-5

       = 4.75

PH = 4

PH     = Pka + log[CH3COONa]/[CH3COOH]

   4       = 4.75 + logx/0.1

logx/0.1    = 4-4.75

logx/0.1    = - 0.75

   x/0.1     = 10^-0.75

   x/0.1     = 0.1778

   x          = 0.1778*0.1

x             = 0.01778

no of moles of CH3COONa = 0.01778moles

mass of CH3COONa = no of moles * gram molar mass

                                 = 0.01778*82   = 1.458g