Question

Titrations

 1. A 50.00-ml NaOH sample of unknown concentration was titrated with 0.1274 M HCI. if 33.61 mL of the HCl solution were required to neutralize the NaOH sample, what is the molarity of the NaOH sample? Show the steps in your calculation 

 2. A 25.00-ml HCl sample of unknown concentration was titrated with 0.5631 M Al(OH)₃. If 37.62 mL of the Al(OH) solution were required to neutralize the HCl sample, what is the molarity of the HCl sample? (Assume Al(OH)₃ is acting as a strong base)

Answer 1

From the condition of equivalency we know,

where C is the molar concentrations of H⁺ (or actual acid) or, OH^- (or actual base).

1. Here, V_acid = 33.61 mL , C_acid = 0.1274 M (because H⁺ is monoprotic) and V_base = 50.00 mL

Thus,

So, the base concentration is 0.0856 M.

2. Here, V_base = 37.62 mL, C_base = 3 x 0.5631 M =1.6893 M (because actual base is OH^- and it is supplying 3 moles of OH^- per mole of Al(OH)₃ ), V_acid = 25.00 mL

Thus,

So, acid concentration is 2.5420 M.