1a. determine the pH of a citrate buffer that is 0.14 M citric acid (H3C6H5O7) a
ID: 967759 • Letter: 1
Question
1a. determine the pH of a citrate buffer that is 0.14 M citric acid (H3C6H5O7) and 0.18 M sodium citrate (NaH2C6H5O7). (What if it were 0.14 sodium citrate and 0.18 M disodium citrate (Na2HC6H5O7)?
2a. For the origional citrate buffer in question 1a, determine what the pH will be if 10mL of 1.09 M HCl solution is added to 250mL of the buffer?
3a. For the origional citrate buffer in wuestion 1a, determine how much 1.09 M HCl could beadded before you excedd the effective capacity of the buffer. (How much 1.09 M HCl would be required to get [A-]/[HA} to equal 0.1)?
Explanation / Answer
1(a)
H3C6H5O7 is the acid and its molarity is given as o.14M, and NaH2C6H5O7 is the conjugate base with molarity 0.18M. Here we need Pka1 for the acid and it is 3.15.
From handerson equation, pH = Pka + log(base/acid)
pH = 3.15 + log(0.18/0.14)
pH = 3.15 + 0.109
pH = 3.259
1(b)
Now if the concentrations are opposite what we have in (a) then...
pH = 3.15 + log(0.14/0.18)
pH = 3.15 -0.109
pH = 3.041
2(a) initial moles of H3C6H5O7 = 0.14 * 0.250 = 0.035
initial moles of NaH2C6H5O7 = 0.18 * 0.250 = 0.045
moles of HCl added = 1.09 * 0.01 = 0.0109
being an acid, HCl would react with NaH2C6H5O7 to form weak acid, H3C6H5O7.
So, new moles of H3C6H5O7 = 0.035 + 0.0109 = 0.0459
and new moles of NaH2C6H5O7 = 0.045 - 0.0109 = 0.0341
pH = 3.15 + log(0.0341/0.0459)
pH = 3.15 - 0.129
pH = 3.021
3(a)
Given that, [A-]/[HA] = 0.1
pH = 3.15 + log(0.1)
pH = 3.15 - 1
pH = 2.15
let's say X moles of HCl were added.
let's assume 1 L of buffer.
initial moles of HA = 0.14
initial moles of A- = 0.18
final moles of HA = 0.14 + X
final moles of A- = 0.18 - X
2.15 = 3.15 + log[(0.18 - X)/(0.14 + X)
2.15 - 3.15 = log[(0.18 - X)/(0.14 + X)
-1 = log[(0.18 - X)/(0.14 + X)
10^-1 = [(0.18 - X)/(0.14 + X)
0.1 = [(0.18 - X)/(0.14 + X)
0.1(0.14 + X) = (0.18 - X)
0.014 + 0.1X = 0.18 - X
0.1X + X = 0.18 - 0.014
1.1X = 0.166
X = 0.166/1.1 = 0.151
so, moles of HCl added are 0.151
molarity of HCl is 1.09M
so, Liters of HCl added = 0.151/1.09 = 0.139
so, we need to add 0.139 L of 139 ml of HCl.
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