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Question 5 A. How many mL of 0.1 M CH3COONa must you add to 35.0 mL...

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 H3COONa 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 H3COONa. How many millimoles of H3COONa should you add to this solution?

Expert Solution

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

= 0.5*0.035 = 0.0175

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

4.7 = 4.75 + log[CH3COONa]/CH3COOH]

log[CH3COONa]/CH3COOH] = 4.7-4.75

log[CH3COONa]/CH3COOH] = -0.05

[CH3COONa]/CH3COOH] = 10^-0.05

[CH3COONa]/CH3COOH] = 0.89

[CH3COONa] = 0.89*[CH3COOH]

= 0.89*0.0175

= 0.015575 moles

no of moles = molarity * volume in L

0.015575 = 0.1* volume in L

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

B. POH = 14-PH

= 14-8.8 = 5.2

PKb = 14-PKa

= 14-8.08 = 5.92

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/base-x = 0.19

x/0.01-x = 0.19

x = 0.19*(0.01-x)

x = 0.00159

no of moles = molarity * volume in L

0.00159 = 0.1* volume in L

volume in L = 0.0159L = 159ml >>>> answer

c. no of moles 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]

= 4.75 + log0.0035/0.0015

= 4.75 + 0.3679 = 5.1179

d. PH = PKa + log[CH3COONa]/[CH3COOH]

4.68 = 4.75 + log15/[CH3COOH]

log15/[CH3COOH] = 4.68-4.75

log15/[CH3COOH] = -0.07

15/[CH3COOH] = 10^-0.07

15/[CH3COOH] = 0.85

[CH3COOH] = 15/0.85 = 17.64 millimoles

queen_honey_blossom answered 1 year ago

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