Question

Titration of 25.00 mL of a 7.35*10^-2M Phosphoric acid solution with a secondary NaOH solution with concentration of .1225M. What is the pH after the following addition of NaOH?

a. 10mL

b. 15mL

Answer 1

a)

mmol of acid = MV = 25*(7.35*10^-2) = 1.8375

mmol of base added = MV = 10*0.1225 = 1.225

the neutralization

H3PO4 + NaOH = H2PO4- + Na+ + H2O

mmol of H3PO4 left = 1.8375 -1.225 = 0.6125

mmol of H2PO4- formed = 0+1.225 = 1.225

this is a buffer...

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

Now,

pH = pKa + log(H2PO4- /H3PO4)

pH = 2.12 + log(1.225 /0.6125)

pH = 2.42

b)

V = 15 mL

mmol of acid = MV = 25*(7.35*10^-2) = 1.8375

mmol of base added = MV = 15*0.1225 = 1.8375

this is the FIRST neutralization of H3PO4 proton

H3PO4 + NAOH = H2PO4- + H2O

there is NOI H3PO4 left

only H2PO4- present

the pH is calculated

ph = 1/2*(pKa1 + pKA2)

since pKa1 and pKa2 are fixed, the equivalence point is exactly in the middle between these pointds

pH = 1/2*(2.12 + 7.21)

pH = 4.665