1.) If 16.45 mL of 0.135 M nitric acid neutralizes 26.0 mL of ammonium hydroxide, what is the molarity of the base?

HNO3(aq) + NH4OH(aq) --> NH4NO3(aq) + H2O(l)

I used stoichiometry to find the # of moles of ammonium hydroxide:

(0.01645 L HNO3) x (0.135 mol HNO3/1 L HNO3) x (1 mol NH4OH/1 mol HNO3)= 0.002221 mol NH4OH

Then used the molarity formula with the volume for the base:

0.002221 mol NH4OH/0.026 L NH4OH= 0.0854 M

Is this the correct answer (0.0854 M)?

2.) If the titration of a 24.0 ml sample of acetic acid requires 25.90 ml of 0.010 M calcium hydroxide, what is the molarity of the acid?

2HC2H3O2(aq) + Ca(OH)2(aq) --> Ca(C2H3O2)2(aq) + 2H2O(l)

I did this one the same way by using stoichiometry first to find moles of acetic acid:

0.02590 L Ca(OH)2 x (0.010 mol Ca(OH)2/1 L Ca(OH)2) x (2 mol HC2H3O2/1 mol Ca(OH)2)= 0.000518 mol acid

Again, I used the molarity formula with the volume given for acetic acid in the problem:

0.000518 mol HC2H3O2/ 0.024 L= 0.0216 M

Is this ans werrect answer (0.0216 M)?

Your time and help are greatly appreciated by me! Thank you so much!

## Comments

Part 2 should be 00054 M. 2 mole acetic acid belongs in denominator. Sorry. I followed your calculations without writing it out. Thinking about it bothered me and I wroted it out myself.

Thank you very much for reviewing my problems George!

If I put the 2 moles of acetic acid in the denominator, my moles of calcium hydroxide won't cancel...right? Maybe I am just looking at it wrong...

Hi Victoria,

You were right with the work you originally posted. The 2 mol of acetic acid should be in the numerator as you have it. Otherwise, you are correct that the moles of calcium hydroxide won't cancel. Good job!

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