Question 5 A. How many mL of 0.1 M CH3COONa must you add to 35.0 mL of 0.50 M CH
ID: 1067788 • Letter: Q
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 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?
Explanation / Answer
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
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