Question

The total acid concentration (nitric+nitrousacid) of the10.0mL solution of “acid rain” is determined by an acid-base titration method. You add 5.00 mL of the titrant sodium hydroxide with a concentration of 0.0114 M to reach the endpoint.
i. What is the number of moles of hydroxide ions added?
ii. Write the net ionic equation for the acid-base reaction.
iii. What is the total number of moles of hydrogen ions in the “acid rain” sample?
iv. What is the total concentration of hydrogen ions in the “acid rain” sample?

Answer 1

(i)

Molarity = Moles / Liter

Moles = Molarity x Liter

5.00 mL of 0.0114 M sodium hydroxide

So, moles of NaOH = 0.0114 M x 0.005 L = 5.7 x 10^-5 moles

(ii)

Since its a reaction of a strong acid (nitric acid + nitrous acid) and strong base (NaOH). So, the net ionic equation for the acid-base reaction is'

H⁺ + OH^-   H₂O

(iii)

moles of NaOH = 0.0114 M x 0.005 L = 5.7 x 10^-5 moles

So, moles of OH^- = 0.0114 M x 0.005 L = 5.7 x 10^-5 moles.

Hence the moles of hydrogen ions = 5.7 x 10^-5 moles.

(iv)

Moles of hydrogen ions = 5.7 x 10^-5 moles

Total volume = 10 mL + 5 mL = 15 mL = 0.015 L

So,

[H⁺] = (5.7 x 10^-5 moles) / 0.015 L = 0.0038 M