Question

Calculate the pH at the equivalence point in titrating 0.110 M solutions of each of the following with 9.0×10⁻²MNaOH.

Answer 1

NaOH is base we then used acid for titration .Let acid is HA .

NaOH +HA---> H2O + NaA

1mol      1mol     1mol

mol of acid = 0.11     [ let volume is =1L]

mol of NaOH = 0.09

0.11mol acid + 0.09 mol NaOH = 0.09 mol H2O + 0.02mol H+

at equvalence point mol of acid =mol of base

after addition of acid & base

molarity of acid = 0.11/2 = 0.055M

Ka for = z

Ka = [H+][CA-] /[HA]

z =x^2/ 0.055-x

z - zx - x^2 = 0

x= [H+]

pH = -log[H+]=

The pH at the equivalence point depends on the concentration of the conjugate base of the acid and the strength of the conjugate base.

for strong acid pH at equivalence point = 7

since the conjugate base is very weak .

-    For other the acid solutions , calculate the volume of NaOH solution was added. Also calculate the number of moles of acid calculated and divide by the total volume (volume of the acid + the volume of the base). This will give you the concentration of the conjugate base.

Kw = Kb x Ka

Kb = Kw/Ka

Kb = [HA] [OH-]/[A-]

this way we get [OH-] & so [H+]

then we can calculate pH of resulting solution.