Consider a 500 mL buffer made of 0.500 M formic acid (Pk. = 3.75) and 0.250 M sodium formate. a. What is the initial pH of the buffer? b. What is the change in pH after 50 mL of 1.00 M HCI has been added to the buffer? c. If the original buffer (from part a) had either 0.600 M HCl or 0.400 M NaOH spilled into it an the new pH was measured to be 4.12; which was spilled and how many mLs?

Fundamentals Of Analytical Chemistry
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Chapter14: Principles Of Neutralization Titrations
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Consider a 500 mL buffer made of 0.500 M formic acid (Pk₂= 3.75) and 0.250 M sodium formate.
a. What is the initial pH of the buffer?
b. What is the change in pH after 50 mL of 1.00 M HCI has been added to the buffer?
c.
If the original buffer (from part a) had either 0.600 M HCl or 0.400 M NaOH spilled into it and
the new pH was measured to be 4.12; which was spilled and how many mLs?
a. pH= pka + loy (HA)
lug (0.250)
= 3.75 lug
= 4.05
b) pH = pka + lug (HA=X)
= 3.75 + log (0.1)
= 3.81
c) pH= pka + log (=
HA +X
0.250-0.0501
+
NHA CV = (0.500 M) (0.500L)
= 0.250 mol
12.5 = 0.250 +X
0.125-X
157 -12.5x=0.250 A
I5x = 182
nA = CV = (0.250M) (0.50UL)
= 0.125 mul
0.250+
x
4.12 = 3.75 + lug (01259 - X
1.09 = log (0.250+)
X = 0.158 mol
X = nHC₁ = CV = (1.00m) (0.050L)
= 0.050 mol
pH 9, and spilled
Vact= пус
= 0.158 mol x0.600M
= 0.0948 L
Transcribed Image Text:Consider a 500 mL buffer made of 0.500 M formic acid (Pk₂= 3.75) and 0.250 M sodium formate. a. What is the initial pH of the buffer? b. What is the change in pH after 50 mL of 1.00 M HCI has been added to the buffer? c. If the original buffer (from part a) had either 0.600 M HCl or 0.400 M NaOH spilled into it and the new pH was measured to be 4.12; which was spilled and how many mLs? a. pH= pka + loy (HA) lug (0.250) = 3.75 lug = 4.05 b) pH = pka + lug (HA=X) = 3.75 + log (0.1) = 3.81 c) pH= pka + log (= HA +X 0.250-0.0501 + NHA CV = (0.500 M) (0.500L) = 0.250 mol 12.5 = 0.250 +X 0.125-X 157 -12.5x=0.250 A I5x = 182 nA = CV = (0.250M) (0.50UL) = 0.125 mul 0.250+ x 4.12 = 3.75 + lug (01259 - X 1.09 = log (0.250+) X = 0.158 mol X = nHC₁ = CV = (1.00m) (0.050L) = 0.050 mol pH 9, and spilled Vact= пус = 0.158 mol x0.600M = 0.0948 L
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