What is the Hydroxide Ion Concentration in an Ionic Equilibrium Problem?

[Speedking96]

Homework Statement

Calculate the hydroxide ion concentration:

16.5 mL of aqueous sulfuric acid at 1.5 M added to 12.7 mL of sodium hydroxide at 5.5M.

2. The attempt at a solution

H₂SO_{4 (aq)} + 2 NaOH _(aq) <--> 2 H₂O _(l) + Na₂SO_{4 (aq)}

Hydrogen Concentration: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
Hydroxide Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles

0.0451 moles of Hydroxide / (16.5 + 12.7 mL) = 1.54 M

 

[Borek]

This is hardly an equilibrium, looks like a simple stoichiometry to me.

You have wrote the reaction equation (good idea) but then you ignored it (bad idea).

 

[Speedking96]

H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

Hydrogen Concentration: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
Hydroxide Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles x 2 = 0.1397

0.0.11495 moles of Hydroxide / (16.5 + 12.7 mL) = 3.93 M

 

[Borek]

With some luck third guess will be the correct one.

 

[Speedking96]

Why? I incorporated the NaOH coefficient.

 

[Borek]

Just because you multiplied something by 2 doesn't mean you did it correctly.

How many moles of acid do you have? How many moles of NaOH will react with this acid?

 

[Speedking96]

H2SO4 (aq) + 2 NaOH (aq) <--> 2 H2O (l) + Na2SO4 (aq)

Acid: (16.5 mL / 1000) (1.5 M) = 0.02475 moles
NaOH Concentration: (12.7 mL / 1000) (5.5 M) = 0.06985 moles

0.0495 moles of NaOH will react with the acid.

Extra OH: 0.06985 - 0.0495 = 0.02035 moles

Concentration: (0.02035) / (16.5 + 12.7 mL) = 0.7 M

 

[Borek]

Looks OK now.