Question

Fifty (50.00) milliliters of 0.02000M 2-(N-morpholino)ethanesulfonic acid (pKa= 6.27) is titrated with 0.1000 M NaOH. Calculate the pH before any base is added, at the half-equivalence point, just before the equivalence point, (d)at the equivalence point, and 20 mL past the equivalence point.

Answer 1

- VNaOH = 0 mL

HA + H2O = A- + H3O-

Ka = 10 ^ -pKa = 10 ^ -6.27 = 5.37x10 ^ -7

In the balance we have:

Ka = [A-] * [H3O +] / [HA] = X ^ 2 / 0.02 - X

By clearing and applying the second degree equation, we have:

X = [H3O +] = 1.03x10 ^ -4

We calculate pH = -Log (1.03x10 ^ -4) = 4

- VNaOH = Mean point of equivalence

pH = pKa = 6.27

- VNaOH = Nearly equivalent point:

V NaOH eq = Ca * Va / Cb = 0.02 * 50 / 0.1 = 10 mL

V NaOH to be used = 9 mL

Total V = 59 mL

calculating the moles of added base:

n NaOH = 0.1 mol / L * 0.009 L = 0.0009 moles = n A- formed

calculating the initial moles of acid:

n HA = 0.02 * 0.05 = 0.001 moles

calculating the final moles of acid:

n Final HA = 0.001 - 0.0009 = 0.0001 mol

calculating the concentrations:

[A-] = 0.0009 / 0.059 = 0.015 M

[HA] = 0.0001 / 0.059 = 0.0017 M

calculating pH by henderson equation - hasselbalch:

pH = pKa + Log [A -] / [HA] = 6.27 + Log (0.015 / 0.0017) = 7.22

- VNaOH = 10 mL

Vt = 60 mL

reaction that occurs:

A- + H2O = HA + OH-

we calculate concentration of A-:

[A-] = 0.001 mol / 0.06 = 0.017

calculating Kb = Kw / Ka = 10 ^ -14 / 5.37x10 ^ -7 = 1.86x10 ^ -8

In the balance we have:

Kb = X ^ 2 / 0.017 - X = 1.86x10 ^ -8

Applying clearance and equation of the second degree we have:

X = [OH-] = 1.78x10 ^ -5

pOH = -Log (1.78x10 ^ -5) = 4.75

pH = 14 - 4.75 = 9.25

- VNaOH = 30 mL

V t = 80 mL

We calculate the moles of NaOH added:

n NaOH = 0.1 * 0.03 = 0.003 mol

n Acid = 0.001

n NaOH f = 0.003 - 0.001 = 0.002 mol

[NaOH] f = 0.002 / 0.08 = 0.025

pOH = -Log (0.025) = 1.60

pH = 14 - 1.6 = 12.4.