Question

You decide to perform a titration using 15 mL of a 1.2 M NH3 as your analyte and 0.6 M HCl as your titrant. Assume a new 15 mL of analyte is used for each step, determine the pH when the following volumes of titrant are used. (Kb =1.8x10-5)

(a)0 mL titrant added.

(b)20 mL titrant added

(c)30 mL titrant added

(d)40 mL titrant added

How do you get the answers :

a. (weak base) pH = 11.678

b. (buffer) pH = 8.968

c. (conjugate weak acid) pH = 4.838

d. (xs strong acid) pH = 0.962

Please show work I am confused

Answer 1

a) At 0 mL of titrant the reaction occurs:

NH3 + H2O = NH4 + + OH-

From the expression of Kb:

Kb = [NH4 +] * [OH-] / [NH3]

1.8x10 ^ -5 = X ^ 2 / 1.2

It clears X = 0.0046 M

The pOH and pH are calculated:

pOH = - log 0.0046 = 2.33

pH = 14 - 2.33 = 11.67

b) The initial moles of NH3 and those of HCl added are calculated:

n NH3 = M * V = 1.2 M * 0.015 L = 0.018 mol

n HCl = 0.6 M * 0.02 L = 0.012 mol

The HCl reacts with the NH3 and forms NH4 +, the pH is calculated:

pH = - log Ka + log (n NH3 / n NH4 +) = - log 5.6x10 ^ -10 + log (0.018 - 0.012 / 0.012) = 8.95

c) At the equivalence point the reaction occurs:

NH4 + + H2O = H3O + + NH3

The concentration of NH4 + formed in the 45 mL solution is calculated:

[NH4 +] = 1.2 M * 15 mL / 45 mL = 0.4 M

You have the expression of Ka:

Ka = [H3O +] * [NH3] / [NH4 +]

5.6x10 ^ -10 = X ^ 2 / 0.4

It clears X = 1.5x10 ^ -5 M

The pH is calculated:

pH = - log 1.5x10 ^ -5 = 4.82

d) The moles of HCl added are calculated:

n HCl = 0.6 M * 0.04 L = 0.024 mol

The remaining moles HCl and its concentration in 55 mL are calculated:

n HCl = 0.024 - 0.018 = 0.006 mol

[HCl] = n / V = ​​0.006 / 0.055 = 0.11 M

The pH is calculated:

pH = - log 0.11 = 0.962

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.