Please help! I know the answer but do not know how to get it. Citric acid (H 3 C
ID: 894874 • Letter: P
Question
Please help! I know the answer but do not know how to get it.
Citric acid (H3C6H5O7) is a natural preservative found in citrus fruits. It is also used as an additive to add an acidic or sour taste to foods and beverages. In addition to its many uses, it can also be utilized to create a buffer to study mammalian tissues. A scientist has a 500 mL solution of 1.0 M H3C6H5O7 / 1.0 M NaH2C6H5O7. She then adds in 400 mL of 0.50 M KOH. What is the pH after the addition of KOH at 25°C?
pKa1 = 3.13 pKa2 = 4.76 pKa3 = 6.40
A. 3.50 <--- Correct Answer
B. 4.39
C. 5.13
D. 2.76
E. 7.38
Explanation / Answer
H3C6H5O7
V = 500 ml
M acid = 1 M of H3C6H5O7
M conjugate = 1 M NaH2C6H5O7
V2 = 400 ml
M2 = 0.5 KOH
a) Find pH
We need pKa values, but first, let us see what is the equilibrium
H3C6H5O7 <---> H+(aq) + H2C6H5O7+(aq)
NaH2C6H5O7 --> Na+(aq) + H2C6H5O7+(aq)
There is a common ion in the first ionization!
Therefore, use pka1
pKa1 = 3.13
This is a buffer, is best described with Hnederson Hasslelbach equation
pH = pKa + log(A-/HA-)
pH = 3.31 + log(0.1/0.1) = 3.13
the original pH is 3.13
Now let us find the pH when adding NaOH
find moles of NaOh
moles = M*V = 0.4L*0.5M =0.20 mol of NaOH
Find out how many moles of acid are present
moles = M*V = 0.5L*1 = 0.5 mol of Acid
H3C6H5O7 + NaOH --> H2C6H5O7+(aq) + H2O(l) + Na(aq)
Note that this neutralization will take out acid, and produce more conjugate...
Lets find out how many
Moles of Acid reacted = 0.5-0.2 = 0.30 left
Moles of Conjugate formed = 0.5 + 0.2 = 0.7
Susbtitute in pH equation
pH = pKa + log(A-/HA-) = 3.13+ log(0.7/0.30) = 3.13 + 0.368 = 3.497 or 3.50
pH = 3.50