Answer:
a) We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2
b) We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2
c) We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2
Explanation:
a) <em>What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = pKa?</em>
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Step 1: Data given
Volume of HC2H3O2 = 1.0 L
Molarity of HC2H3O2 = 1.8 M
Ka = 1.8*10^-5
ph = pK = -log(1.8*10^-5) = 4.74
Step 2:
Use the Henderson-Hasselbalch equation.
pH = pKa + log(A-/HA)
4.74 = 4.74 + log(A-/HA)
0 = log(A-/HA)
A-/HA = 1
Consider X = moles of NaOH added (and moles of A- formed)
Remaining moles of HA = 1.8 - X
moles of A- = X
HA = 1.8 - X
X/(1.8-X) = 1
X =0.9
<u>We have to add 0.9 mole NaOH to 1.0 L of 1.8 M HC2H3O2
</u>
To control we can do the following equation:
4.74 = 4.74 + log(0.9/0.9) = 4.74
b)<em> What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 4.00?</em>
Step 1: Data given
Volume of HC2H3O2 = 1.0 L
Molarity of HC2H3O2 = 1.8 M
Ka = 1.8*10^-5
ph = 4
Step 2:
Use the Henderson-Hasselbalch equation.
pH = pKa + log(A-/HA)
4 = 4.74 + log(A-/HA)
-0.74 = log(A-/HA)
A-/HA = 0.182
Consider X = moles of NaOH added (and moles of A- formed)
Remaining moles of HA = 1.8 - X
moles of A- = X
HA = 1.8 - X
X/(1.8-X) = 0.182
X =0.277
<u>We have to add 0.277 mole NaOH to 1.0 L of 1.8 M HC2H3O2
</u>
<u>
</u>
To control we can do the following equation:
4 = 4.74 + log(0.277/1.523)
<em></em>
<em>c) What quantity (moles) of NaOH must be added to 1.0 L of 1.8 M HC2H3O2 to produce a solution buffered at pH = 5.00</em>
Step 1: Data given
Volume of HC2H3O2 = 1.0 L
Molarity of HC2H3O2 = 1.8 M
Ka = 1.8*10^-5
ph = 5
Step 2:
Use the Henderson-Hasselbalch equation.
pH = pKa + log(A-/HA)
5 = 4.74 + log(A-/HA)
0.26 = log(A-/HA)
A-/HA = 1.82
Consider X = moles of NaOH added (and moles of A- formed)
Remaining moles of HA = 1.8 - X
moles of A- = X
HA = 1.8 - X
X/(1.8-X) = 1.82
X =1.16
<u>We have to add 1.16 mole NaOH to 1.0 L of 1.8 M HC2H3O2
</u>
<u>
</u>
To control we can do the following equation:
5 = 4.74 + log(1.16/0.64) = 5
