Titration question

Question

In the titration of a solution of Sodium hydroxide, an acid titrant was prepared by diluting 125.0 mL of 10.00 mol/L Nitric acid into enough distilled water to make 500.0 mL of solution. 30.0 mL of the base was measured, and in the titration, 16.74 mL of the acide titrant was needed to neutralize. Determine the concentration of the Sodium hydroxide.

Answer 1

Hello Lauren

 

I tried to add a comment here regarding the following, but the website was not displaying it.

 

I just wanted to point out that in Steps #3 and #5 of your answer, the volume of NaOH should be 30.0 + 16.74 = 46.7 mL (rounded to 1 decimal) since the volumes of the acid and base are additive in a titration.

 

Thank you!

Answer 2

.125 L HNO3 x 10 mole HNO3/ L HNO3 = 1.25 moles HNO3

 

Conc x vol = conc x vol

 

16.74 mL x 1.25 mole HNO3/500mL = 30 mL NaOH x [ mole NaOH/mL]

 

16.74 mL x 1.25 mole/500mL = .04185 mol HNO3

 

.04185 mol HNO3 = 30 mL NaOH x [ mole NaOH/mL]

 

.04185 mol HNO3/30 mL = .001395 mol of NaOH/mL

 

.001395 mol of NaOH/mL x 1000mL/L = 1.395mol NaOH/L