Question 5 a) How many mL of 0.1 M CH3COONa must you add to 35.0 mL of 0.50 M CH

Question

Question 5

a) How many mL of 0.1 M CH3COONa must you add to 35.0 mL of 0.50 M CH3COOH to make a buffer of pH 4.7? [pKa of H3COONa is 4.75]

b) Use the Henderson-Hasselbalch equation to calculate how many mL of 0.1 M HCl must you add to 100 mL of 0.10 M Tris base to make a buffer of pH 8.8? [pka of Tris Base is 8.08]

c) What is the pH when 25.0 mL of 0.200 M of CH3COOH has been titrated with 35.0 mL of 0.100 M NaOH? Write out the balanced chemical equation.

d) *You need to produce a buffer solution that has a pH of 4.68. You already have solution that contains 15.0 mmol of CH3COOH. How many millimoles of CH3COONa should you add to this solution?

Explanation / Answer

no of moles of CH3COOH   =molarity * volume in L

                                            = 0.5*0.035   = 0.0175 moles

PH   = PKa + log[CH3COONa]/[CH3COOH]

4.7   = 4.75 + log[CH3COONa]/[CH3COOH]

4.7-4.75     = log[CH3COONa]/[CH3COOH]

log[CH3COONa]/[CH3COOH] = -0.05

[CH3COONa]/[CH3COOH]       = 10-0.05

[CH3COONa]/[CH3COOH]      = 0.89

[CH3COONa]                         = 0.89*[CH3COOH]

[CH3COONa]                       = 0.89*0.0175

no of moles of CH3COONa = 0.015575moles

no of moles = molarity * volume in L

0.015575     = 0.1* volume in L

volume in L   = 0.015575/0.1 = 0.15575 L   = 155.75 ml >>> answer

b. no of moles of Tris base = molarity * volume in L

                                            = 0.1*0.1 = 0.01 moles

   POH   = 14-PH

             = 14-8.8   = 5.2

   POH   = Pkb + log[Salt]/[base]

   5.2 = 5.92 + logx/[base-x]

logx/[base-x]   = 5.2-5.92

logx/[base-x]    = -0.72

x/base-x        = 10-0.72

x/0.01-x       = 0.19

x                  = 0.19*(0.01-x)

x                  = 0.0159

no of moles of HCl = 0.0159moles

no of moles   = molarity * volume in L

0.0159           = 0.1*volume in L

volume in L    = 0.0159/0.1   = 0.159L = 159ml

c.              no of moles of CH3COOH = molarity * volume in L

                                                             = 0.2*0.025   =0.005moles

             no of moles of NaOH             = molarity * volume in L

                                                            = 0.1*0.035   = 0.0035 moles

                CH3COOH + NaOH ----------> CH3COONa + H2O

   I         0.005             0.0035                    0

   C     -0.0035           -0.0035                  0.0035

   E     0.0015             0                           0.0035

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

            PH     = 4.75 + log0.0035/0.0015

                      = 4.75 + 0.3679 = 5.1179

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