Question

Sketch a titration curve for 50 mL of 0.10 M HCl titrated with 0.15 M NaOH.

Include on the graph:

a) The initial pH

b) The volume of the equivalence point

c) The pH at the equivalence point

d) The pH after addition of 50 mL of NaOH

SHOW STEPS

Answer 1

1. HCl isa strong acid and NaOH is a strong base, thus both of them completely dissociate in aqueous solution. Hence, conc. of H+ ions is equal to conc. of HCl and Conc. of OH- ions is equal to conc. of NaOH. The initial pH of the solution containing 50mL of 0.10M HClcan becalculated as follows:
   
   [H+] = Concentration of HCl (mol/dm3)
   [H+] = 0.10 mol/dm³
   Thus, pH = -log[H+]= -log(0.10)
   pH = 1.0
   Therefore, initial pH of the solution is 1.0

2. At equivalence point moles of NaOH will be equal to moles ofHCl. Thus, moles of HCl in given 50mL solution can be determined as follows:
   Moles of HCl = Concentration of HCl (mol/dm3) X volume of HCl
   Moles of HCl = 0.10 mol/dm3 x 0.05dm3
   Moles of HCl = 0.005 mol
   
   Thus, moles of NaOH will also be 0.005mol at equivalence point. Volume of NaOH can be calculated as follows:
   
   Volume of NaOH = Moles ofNaOH / concentration(mol/dm3)
   Volume of NaOH = (0.005 mol) / (0.15mol/dm3)
   Volume of NaOH = (0.005 mol) / (0.15mol/dm3)
   Volume of NaOH = 0.033dm³
   
   Therefore, at equivalence point volume of 0.15M NaOH will be 33mL or 0.033dm³

3. pH at equivalencepoint willbe pH of water as there will not be any H+ or OH- ions. Thus, at equivalence point pH will be 7.0

4. After addition of 50mL of NaOH, amount of NaOH remaining after neutralization of HCl will be 50mL – 33mL = 17mL.
   Concentration of OH- ions in 17mL of 0.15M NaOH can be calculated as follows:
   [OH-] = concentration of NaOHin mol/dm3
   [OH-] = 0.15 mol/dm3
   
   pOH = -log[OH]
   pOH = -log(0.15)
   pOH = 0.82
   
   Thus, pH = 14-pOH = 14-0.82 = 13.18
   Therefore, pH after addition of 50mL of NaOH will be 13.18

5. pH after addition of 5mL of 0.15M NaOH solution can be determined as follows:
   Moles of NaOH in 5mL of 0.15M NaOH can be calculated as follows:
   Moles of NaOH = (0.15 mol/dm3) x 0.005mL
   Moles of NaOH = 0.00075mol
   
   Thus, moles of HClremaining after reacting will 5mL of 0.15M NaOH= 0.005 – 0.00075 = 0.00425moles
   Total volume of solution= volume of HCl + volume of NaOH = 50mL + 5mL = 55mL
   
   Thus, [H+]= 0.00425mol/0.055dm3
   [H+] = 0.077
   
   Thus, pH = -log(0.077) = 1.11
   
   Therefore, after addition of 5mL of 0.15M NaOH, pH of the solution changes to 1.11
   
   Similarly, pH after addition of each 5mL of NaOH solution can be calculated. Calculated values are shown in table given below:
   

   Sr. No.	mL of NaOH added	pH
   1	0	1.0
   2	5	1.11
   3	10	1.24
   4	15	1.38
   5	20	1.55
   6	25	1.79
   7	30	2.21
   8	35	11.47
   9	40	12.04
   10	45	12.25
   11	50	13.18