# Homework Help: Ionic Equilibrium Problem

1. Jun 7, 2014

### Speedking96

1. The problem statement, all variables and given/known data

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

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

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

2. Jun 8, 2014

### Staff: Mentor

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).

3. Jun 8, 2014

### 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

4. Jun 8, 2014

### Staff: Mentor

With some luck third guess will be the correct one.

5. Jun 8, 2014

### Speedking96

Why? I incorporated the NaOH coefficient.

6. Jun 8, 2014

### Staff: Mentor

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?

7. Jun 8, 2014

### 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

8. Jun 8, 2014

### Staff: Mentor

Looks OK now.