1. A beaker with 2.00×102 mL of an acetic acid buffer with a pH of 5.000 is sitt
ID: 538683 • Letter: 1
Question
1. A beaker with 2.00×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 7.90 mL of a 0.490 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740. find pH =
2. When species combine to produce a coordination complex, the equilibrium constant for the reaction is called is the formation constant, Kf.
For example, the iron(II) ion, Fe2+, can combine with the cyanide ion, CN, to form the complex [Fe(CN)6]4 according to the equation
Fe2+(aq)+6CN(aq)[Fe(CN)6]4(aq)
where Kf=4.21×1045.
The average human body contains 5.20 L of blood with a Fe2+ concentration of 2.70×105M . If a person ingests 5.00 mL of 25.0 mM NaCN, what percentage of iron(II) in the blood would be sequestered by the cyanide ion? find % Fe2+
Explanation / Answer
Q1.
V = 2*10^2 mL = 200 mL
acetic acid buffer, p H= 5
total molarity = 0.1 M
total mmolo = MV = 0.1*200 = 20 mmol of acid/conjugate
adding:
V = 7.9 mL of M = 0.49 M HCl acid, mmol of HCl = MV = 7.9*0.490 = 3.871 mmol are added
find pH change
from the buffer equation
ph = pKa + log(A-/HA)
where A- is acetate (conjguaet base) and HA is the acetic acid
substitute known data
A- + HA = 20 mmol
intially
pH = pKa + log(A-/HA)
5 = 4.74 + log(A-/HA)
A-/HA = 10^(5-4.74) = 1.8197
A-/HA =1.8197
and we know
A- + HA = 20 mmol
so
1.8197*HA + HA = 20
HA = 20 / (1.8197+1) = 7.092
then
A- = 1.8197*7.092 = 12.9053
now, after adding
3.871 mmol of HCl added,
mmol of HA formed = 3.871 + 7.092 = 10.963
mmol of A- left = 12.9053 - 3.871 = 9.0343
now, reclaculate pH
pH = pKa + log(a-/HA)
pH = 4.74 + log(9.0343/10.963)
pH = 4.6559
dpH = 4.6559 - 5 = -0.3441
as expected, pH dropped due to HCl addition
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