Question

43.      

In a titration of 35.35 mL of 0.3509 M nitric acid with 0.4242 M aqueous sodium hydroxide, what is the pH of the solution when 35.35 mL of the base have been added?

Answer 1

The neutralization reaction goes like this

HNO₃ + NaOH ===== NaNO₃ + H₂O

let´s calculate the moles of acid with Molarity = moles / volume, moles = Molarity * Volume

moles of Acid = 0.03535 L * 0.3509 M = 0.0124 moles of acid

moles of NaOH = Molarity * volume = 0.03535 * 0.4242 = 0.014995

as we can see there are more moles of base than moles of acid so

moles of NaOH remaining =  0.014995 - 0.0124 = 0.002595 moles of NaOH

Molarity of NaOH remaining = 0.002595 / (0.03535 + 0.03535) = 0.0367 M

since this is comes from the NaOH we will have ions of OH

if we apply -log of OH we will get POH

POH = -log (0.0367) = 1.4352

remember that PH + POH = 14

PH = 14 - POH = 14 - 1.4352 = 12.56 , this is the PH