Question

4. What is the pH of 80 mL of a 0.4 mol/l NaOH solution after the addition 2 points of 20 mL of a 0.5 mol/L solution of HCI? Your answer

Answer 1

NaOH and HCl here will react to form neutral solution of salt and water.

Reaction between NaOH and HCl will be as:
HCl + NaOH ---> NaCl + H₂O

Amount of NaOH = 80ml of 0.4 mol/L
Amount of HCl = 20ml of 0.5 mol/L

Number of moles = Concentration x volume in litres
= Concentration x (volume in mL/1000)

Number of moles of NaOH = 0.4 x (80/1000) = 0.4 x 0.08 = 0.032 mol
Number of moles of HCl = 0.5 x (20/1000) = 0.5 x 0.02 = 0.010 mol

As per the reaction: HCl + NaOH ---> NaCl + H₂O :
1 mol of HCl reacts with 1 mol of NaOH.
0.010 mol of HCl reacts with 0.010 mol of NaOH.

But we have 0.032mol of NaOH therefore some NaOH will be left in the mixtureunreacted. Contribution to pH will be done only by this left out NaOH.

Number of moles of NaOH left = Number of moles of NaOH present initially - Number of moles of NaOH reacted with HCl
= 0.032 - 0.010 = 0.022 mol

Volume of complete mixture = volume of NaOH added + Volume of HCl added = 80ml + 20ml = 100ml = 0.1L

Concentration of NaOH = Number of moles / volume in litres
= 0.022 mol / 0.1 L
= 0.22 mol/L

[OH^- ] = 0.22M

pOH = -log[OH^-] = -log(0.22) = 0.66

and pH + pOH = 14

Therefore, pH = 14- pOH = 14 - 0.66 = 13.34.

Answer = 13.34