25 cm3 of 0.400 mol dm-3 methanoic acid (pKa 3.75) is mixed with 50 cm3 of 0.100 mol dm-3 sodium hydroxide. Calculate the pH of this mixture

Answer 1

HCOOH + NaOH ==> HCOONa + H2O

moles HCOOH = 0.025 L x 0.400 mol/L = 0.01 moles

moles NaOH = 0.05 L x 0.100 mol/L = 0.005 moles NaOH

Final moles HCOOH = 0.01 - 0.005 = 0.005

Final moles HCOONa = 0.005

This can be seen from an ICE table:

HCOOH + NaOH ==> HCOONa + H2O

0.01 ...........0.005................0........................Initial

-0.005........-0.005.............+0.005.................Change

0.005............0...................0.005.................Equilibrium

Now we can use the Henderson Hasselbalch equation, pH = pKa + log [HCOONa]/[HCOOH]

pH = 3.75 + log (0.005/0.005) = 3.75 + log 1

pH = 3.75