Chemistry Study guide

Question

If we titrate 125.0 mL of HCl with 137.0 mL of 3.0 M KOH what is the molarity of the HCl?

HCl (aq) + KOH (aq) → H2O (l) + KCl (aq)

 

Finals are on Monday. I am having trouble understanding how to do this. I know that Molarity=Moles of Solute/Liters of Solvent. But I don't know how to apply that to this question.

Answer 1

For titrations, the quick and dirty way is to use M₁V₁ = M₂V₂, or M_aV_a = M_bV_b  (a = acid and b = base, and if you have everything in mL, you don't even need to convert to L, because if you have the same units on both sides, everything's fine.

If you have a diprotic acid or a diprotic base you can modify the formula to N_aM_aV_a = N_bM_bV_b, where N_a is the number of protons being used in the titration, and N_b is the number of hydroxides being consumed, such as H₂SO₄ being titrated with NaOH (N_a = 2). This can give you a very rapid answer, which may come in handy in a time crunch.

 

For this problem, both being monofuncitonal,  M_aV_a = M_bV_b

 

                                                            M_a x 125 mL = 3.0 M x 137.0 mL

 

                                                             M_a = 3.0 M  x 137.0 mL

                                                                            125.0 mL

 

                                                             M_a = 3.288 M,  which rounds to 3.3 M (2 sig fig's)

Answer 2

First, write the balanced equation so that you can see the mole : mole ratio of reactants.

HCl + KOH ==> KCl + H2O  there is a 1 : 1 mole ratio of HCl : KOH.

How many moles of KOH areused?

   0.1370 L KOH x 3.0 mole/L = 0.411 moles KOH used

How many moles HCl does this represent?

   0.411 moles KOH x 1 mole HCl/mole KOH = 0.411 moles HCl

Molarity of HCl = moles HCl/L = 0.411 moles/0.125 L = 3.288 M (0.125 L is the same as 125.0 ml)

Answer = 3.3 M (to 2 significant figures since 3.0 M has only 2 s.f.)