Question

Calculate the pH for each of the following cases in the titration of 35.0 mL of 0.120 M LiOH(aq), with 0.120 M HCl(aq).

(a) before addition of any HBr (b) after addition of 12.5 mL of HBr (c) after addition of 23.0 mL of HBr (d) after addition of 25.0 mL of HBr (e) after addition of 31.0 mL of HBr

Answer 1

in one cae you written as a HCl and other cases you written as a HBr

ok iam assuming as a HBr and going

Part A before addition of HBr

LiOH is a strong base concentration of OH- = concentration of LiOH

pOH = -log(0.12) = 0.92

pH = 14-pH = 14-0.92 = 13.08

Part-B

lets see the balanced equation

LiOH + HBr ---> LiBr + H2O

means on emole of LiOH will react with on emole of HBr

moles of LiOH = Molarity x volume in liters 0.12 M x 0.035 L = 0.0042 mol

moles of HBr = molarity x volume = 0.12 x 0.0125 L = 0.0015 mol

0.0015mol of HBr will react with 0.0015 moles of LiOH

no of moles of LiOH remaining = 0.0042 - 0.0015 = 0.0027 moles of liOH remaining

total volume = 35+12.5 = 47.5 ml = 0.0475L

concentration of LiOH remaining = 0.0027 mol / 0.0475 = 0.057 M

pOH = -log(0.057) = 1.24

pH = 14-pOH = 14- 1.24 = 12.76

Part-C

moles of LiOH = Molarity x volume in liters 0.12 M x 0.035 L = 0.0042 mol

moles of HBr = molarity x volume = 0.12 x 0.023 L = 0.00276 mol

0.00276mol of HBr will react with 0.00276 moles of LiOH

no of moles of LiOH remaining = 0.0042 - 0.00276 = 0.00144 moles of liOH remaining

total volume = 35+23 = 58 ml = 0.058L

concentration of LiOH remaining = 0.00144 mol / 0.058 = 0.025 M

pOH = -log(0.025) = 1.60

pH = 14-pOH = 14- 1.60 = 12.4

Part-D

moles of LiOH = Molarity x volume in liters 0.12 M x 0.035 L = 0.0042 mol

moles of HBr = molarity x volume = 0.12 x 0.025 L = 0.003 mol

0.003mol of HBr will react with 0.003 moles of LiOH

no of moles of LiOH remaining = 0.0042 - 0.003 = 0.0012 moles of liOH remaining

total volume = 35+25 = 60 ml = 0.06L

concentration of LiOH remaining = 0.0012 mol / 0.06 = 0.02 M

pOH = -log(0.020) = 1.7

pH = 14-pOH = 14- 1.7 = 12.3

Part-e

moles of LiOH = Molarity x volume in liters 0.12 M x 0.035 L = 0.0042 mol

moles of HBr = molarity x volume = 0.12 x 0.031 L = 0.00372 mol

0.00372 mol of HBr will react with 0.00372 moles of LiOH

no of moles of LiOH remaining = 0.0042 - 0.00372 = 0.00048 moles of liOH remaining

total volume = 35+31 = 66 ml = 0.066L

concentration of LiOH remaining = 0.00048 mol / 0.066 = 0.007272 M

pOH = -log(0.007272) = 2.14

pH = 14-pOH = 14- 2.14 = 11.86