Question

10.5 mL of 0.100M ammonium chloride aqueous solution is titrated with 0.100 M NaOH(aq).

What type of solution is present when V_NaOH = 2.00 mL?

Select one:

a. strong acid

b. strong base

c. weak acid

d. weak base

e. buffered

Answer 1

The number of mols of NH₄Cl = (10.5 mL)(0.100 M) = 1.05 mmol

The number of mols of NaOH = (2.00 mL)(0.100 M) = 0.2 mmol

Here, the resultant reaction -

           NH₄⁺ (aq) + OH^- (aq) <--------> NH₃ (aq) + H₂O (l)

         1.05 mmol    0.2 mmol                  0

           -0.2 mmol     -0.2 mmol             +0.2 mmol

          0.85 mmol      0                           0.2 mmol

So the resulting solution will be buffer. So option e) correct.