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)?

Answer 1

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.