Question

Calculate the at the following points in the titration of 20.00 ml of 0.300 M CH 3 COOH with 0.300 M NaOH Ka = 1.8 * 10 ^ - 5

a.Before the addition of any NaOH

b. After the addition of 7.00ml of 0.300 M NaOH

c. After the addition of 10.00 ml of 0.300 M NaOH

d. After the addition of 20.00 ml of 0.300 M NaOH

e. After the addition of 25.00 ml of 0.300 M NaOH

Answer 1

+x Titration of Weak acid with strong base Titration of 20.0 mL of 0.300 M CH3COOH with 0.300 M NaOH answer-a Before addition of NaOH Construct ICE table to obtain their equilibrium concentrations: reaction CH3COOH H H + CH3COO Initial(M) 0.300 0 0 Change(M) -X Equilibrium(M) 0.300-X х Equilibrium expression: Ka = [H+][ CH3COO- 1/[ CH3COOH ] 1.8x10-5 = (x)/( 0.300 -X) Calculating x Value: x + 1.8x10-5 X - 0.00000540 = 0 By solving the above quadratic equation, we get x = H* = 0.002315 M Calculating pH: pH = -log[H+] = -log[ 0.002315 ] = 2.64 X answer-b After addition of 7.00 ml NaOH pKa = -log[ka] = -log[ 1.8x10-5 ] = 4.74 CH3COOH + NaOH H CH3COONa + H2O number of moles - Molarity > Volume number of moles of CH3COOH = 0.300 x 20.0 6.00 mmol number of moles of NaOH 0.300 x 7.0 2.10 mmol moles left of CH3COOH = 6.00 - 2.10 3.90 mmol number of moles of CH3COONa = 2.10 mmol According to Henderson-Hasselbalch equation pH = pka + log[salt/acid] = 4.74 + log[ 2.10 / 3.90 ] pH = 4.48 X

answer-c After addition of 10.00 ml NaOH CH3COOH + NaOH H CH3COONa + H2O number of moles = Molarity Volume number of moles of CH3COOH = 0.300 * 20.0 = 6.00 mmol number of moles of NaOH = 0.300 X 10.0 - 3.00 mmol moles left of CH3COOH = 6.00 - 3.00 3.00 mmol number of moles of CH3COONa = 3.00 mmol According to Henderson-Hasselbalch equation pH = pka + log[salt/acid] = 4.74 + log [ 3.00 1 3.00 ] pH = 4.74 At halfequivalence point pH = pka = 4.74 answer-d After addition of 20.00 ml NaOH CH3COOH + NaOH CH3COONa + H2O number of moles – Molarity > Volume number of moles of CH3COOH = 0.300 x 20.0 = 6.00 mmol number of moles of NaOH = 0.300 x 20.0 6.00 mmol moles left of CH3COOH = 6.00 - 6.00 = 0.00 mmol number of moles of CH3COONa = 6.00 mmol at equivalnece point only salt remains total volume = 20.0 + 20.0 = 40.0 mL moles 6.0 Concentration of CH3COONa = - Volume 40.0 = 0.15 M ( 14 + 4.74 + log 0.15 ] pH = [14+ pKa +logc )/2 = pH = 8.96

x x г answer-e After addition of 25.0 ml NaOH number of moles - Molarity Volume number of moles of CH3COOH = 0.300 20.0 = 6.00 mmol number of moles of NaOH = 0.300 25.0 = 7.50 mmol moles left of NaOH = 7.5 6.0 = 1.50 mmol Total Volume = 20.0 + 25.00 45.0 mL moles 1.50 Concentration of NaOH = = 0.033 M Volume Concentration of OH = 0.033 M pOH = -log[H0] = -log[ 0.033 ] = 1.48 pH = 14 - pOH = 14 - 1.48 = 12.52 + 11 45.0