Answer:
Concentrations of HAc and NaAc you need are 0.122M
Explanation:
pKa of acetic acid is 4.75, that means when amount of sodium acetate and acetic acid is the same, pH will be 4.75
Thus, you know [NaAc]i = [HAc]i
Now, using H-H equation, when pH = 3.75:
3.75 = 4.75 + log [NaAc] / [HAc]
0.1 = [NaAc] / [HAc]
10 [NaAc] = [HAc]
Thus, after the reaction [HAc] must be ten times, [NaAc].
Based in the reaction of NaAc with HCl
NaAc + HCl → HAc + NaCl
Moles of HCl added are:
1mL = 0.001L * (10mol /L) = 0.01 moles HCl.
That means moles of both compounds after the reaction are:
<em>[NaAc] = [NaAc]i - 0.01 mol </em>
[HAc] = [HAc]i + 0.01
Replacing these equations with the information you know:
[NaAc] = [NaAc]i - 0.01 mol
10[NaAc] = [NaAc]i + 0.01
Subtracting both equations:
9[NaAc] = 0.02mol
[NaAc] = 0.0022 moles.
Replacing in <em>[NaAc] = [NaAc]i - 0.01 mol </em>
0.0022mol = [NaAc]i - 0.01 mol
0.0122mol = [NaAc]i = [HAc]i
These moles in 100.00mL = 0.1000L:
[NaAc]i = [HAc]i = 0.0122mol / 0.100L =
0.122M
Thus, <em>concentrations of HAc and NaAc you need are 0.122M</em>
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To create this buffer, you need to pipette 12.2mL of both 1.0M NaAc and 1.0M HAc and dilute this mixture to 100.0mL
