Question

Calculate the pH when 10.00mL of 0.116M NaOH has been added to 25.00mL of 0.132M nitrous acid. pKa for nitrous acid=3.14. Check to see if X can be ignored, if not use the quadratic formula.

Answer 1

Nitrous acid is a weak acid. So, when nitrous acid is added to NaOH, the concentration becomes:

Number of moles of NaOH = Molarity * Volume (in litres) = 0.116*0.010 = 0.00116 moles

Number of moles of nitrous acid = Molarity * Volume (in litres) = 0.132*0.025 = 0.0033 moles

As moles of conjugate base of nitrous acid = moles of NaOH = 0.00166

Moles of nitrous acid unreacted = moles of nitrous acid - moles of NaOH = 0.0033 - 0.00166 = 0.00164

Total volume after mixing = 10+25 = 35 mL or 0.035 L

Molarity of conjugate base after mixing of solutions = Number of moles/ Total volume = 0.00166/0.035 = 0.0474 M

Molarity of nitrous acid after mixing = Left over moles/ Total volume = 0.00164/0.035 = 0.0469 M

Now, as per Henderson Hasselbalch equation,

pH = pKa + log ([conjugate base]/[unreacted acid])

pH = 3.14 + log (0.0474/0.0469)

pH = 3.14 + 0.0048 = 3.145