What is the pH?

Question

Calculate the pH when 50.0 mL of 0.150 M HNO₃ is mixed with 40.0 mL of 0.250 M LiOH.

Answer 1

HNO3 + LiOH ==> LiNO3 + H2O ... balanced equation

mols HNO3 present = 50.0 ml x 1 L / 1000 ml x 0.150 mol/L = 0.0075 moles HNO3

mols LiOH present = 40.0 ml x 1 L / 1000 ml x 0.250 mol/L = 0.010 moles LiOH

Because the react in a 1:1 mole ratio (see balanced equation), this means that all the HNO3 will be reacted leaving some LiOH left over. Since LiOH is a strong base, this will make the pH of the solution >7.

moles LiOH left over = 0.010 mols - 0.0075 mols = 0.0025 mols LiOH

Final volume = 50.0 ml + 40.0 ml = 90.0 mls x 1 L / 1000 ml = 0.090 liters

Final [OH-] = 0.0025 mols OH- / 0.090 L = 0.0278 M

pOH = -log [oh-] = -log 0,0278

pOH = 1.56

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.56

pH = 12.4