Lecture 2
Sample pH problems and solutions

1.  Buffer prepared by mixing 100 ml of 0.01M NaH2PO4 with 100ml 0.01M Na2HPO4.
a. What is pH?

NaH2PO4  =  weak acid = A = 0.001 mols
Na2HPO4  = conjugate base = B = 0.001 mols
pH = pKa + log B/A
= 7.2 + log (0.001/0.001)
= 7.2

b.  What is pH after adding 10 ml of 0.01M HCl?
10ml 0.01M HCl = 0.0001mols
Strong acid converts conjugate base to weak acid.

Therefore: new A = 0.001 + 0.0001 = 0.0011mols
and new B = 0.001 – 0.0001 = 0.0009mols
new pH = 7.2 + log (0.0009/0.0011)
= 7.2 + log(0.82)
=  7.11

c.  What is pH after adding 10 ml of 0.01M NaOH?
10 ml 0.01M NaOH = 0.0001 mols
Strong base converts weak acid to conjugate base.

Therefore new B = 0.001 + 0.0001 = 0.0011 mols
and new A = 0.001 – 0.0001 = 0.0009 mols
New pH = 7.2 + log (0.0011/0.0009)
= 7.2 + log (1.22)
= 7.29
Note that the change in pH (up or down) produced by adding equivalent amounts of strong acid or strong base are equal.  This is only the case when the starting pH of buffer is equal to the  pKa of weak acid.

2. 1.  Buffer prepared by mixing 40ml of 0.01M NaH2PO4 with 100ml 0.01M Na2HPO4.
a. What is pH?

A = 0.0004 mols,  B = 0.001 mols
pH = 7.2 + log(0.001/0.0004)
= 7.6
b.  What is pH after adding 10 ml of 0.01M HCl?
10ml 0.01M HCl = 0.0001mols
Strong acid converts conjugate base to weak acid.

New A = 0.0004 + 0.0001 = 0.0005 mols
New B = 0.001 – 0.0001 = 0.0009 mols
New pH = 7.2 + log (0.0009/0.0005)
=  7.46

c.  What is pH after adding 10 ml of 0.01M NaOH?
10 ml 0.01M NaOH = 0.0001 mols
Strong base converts weak acid to conjugate base.

Therefore new B = 0.001 + 0.0001 = 0.0011 mols
and new A = 0.0004 – 0.0001 = 0.0003 mols
New pH = 7.2 + log (0.0011/0.0003)
= 7.2 + log (3.67)
= 7.76

Note that change in pH produced by adding the same amounts HCl or NaOH are greater in example 2 than in example 1.  In example 1,  pH of buffer is at pKa of weak acid and the buffering capacity is maximum.  Also note that in example 2 the pH changes produced by adding acid and base are no longer equal as was the case in example  1.  This will be true at all pHs except the pKa.  The acid and base buffering capacities are only equal when the pH = pKa.

3.  Given a weak acid,  R-NH3+ with a pKa = 10.4.

a.  What will the net charge be at pH 9.3?
Determine ratio of B/A at pH 9.3.
9.3 = 10.4 + log B/A
-1.1 = log B/A
B/A = .079  and B = .079A
B + A = 1
substituting for B:  1.079A = 1
A = 0.93 and B = 0.07
That is 0.93 or 93% exists in the acidic form.  The net charge is +0.93.

b.  At what pH is the net charge +0.5?
When net charge is =0.5,   50 % of the weak acid exists as R-NH3+ ,  and the remaining 50% exists as R-NH2.     A = B.  This occurs at pKa.  Therefore  pH = 10.4.

c.  At what pH is net charge +0.2.

R-NH3+ = 0.2 = A.  Therefore R-NH2 = 0.8 = B. ( ie R-NH3+ + R-NH2  = 1)

Therefore pH = 10.4 + log (0.8/0.2)
=  10.4 + 0.6= 11.

4.What is pH and percent ionization of a solution of 0.1M acetic acid?  pK acetic acid = 1.74 x 10-5

[H+] = sq rt (Ka x [HA])
= sq rt (1.74 x 10-5 x 0.1)
= 13.2 x 10-4M
and pH =  2.8

% ionization =   [(13.2 x 10-4 ) / 0.1 ] x 100 =  1.32%

2.  What is pH and percent ionization of a solution of 0.01M acetic acid?  pK acetic acid = 1.74 x 10-5

[H+] = sq rt (Ka x [HA])
= sq rt (1.74 x 10-5 x 0.01)
= 4.17 x 10-4M
and pH =  3.38

% ionization =   [(4.17 x 10-4 ) / 0.01 ] x 100 =  4.17%

5.  The pH of a patient's blood was 7.03 and the [CO2] was 1.1 mM.  i.  What was the [HCO3-]?  ii.  How much HCO3-  had
been used in buffering against the acid producing the acidosis?

i.  Using Henderson-Hasselbach:  7.03 = 6.1 + log(HCO3-/CO2)
= 6.1 + log(HCO3-/1.1)
Solving for HCO3-
HCO3- = 9.4 mM
ii.  In normal blood [HCO3-] = 24 mM
Therefore 24 - 9.4 = 14.6 mmols/liter HCO3- had been used to buffer against acid.