Question

Calculate the hypothetical pH AFTER addition of 15.00 mL of reagent for the titration of 50.00 mL of 0.0800 M NaOH with 0.1000 M HCl at 25°C. QUESTION 7 10 poir Calculate the hypothetical pH AFTER addition of 40.00 mL of reagent for the titration of 50.00 mL of 0.0800 M NaOH with 0.1000 M HCl at 25°C.

Answer 1

Calculate the hypothetical pH AFTER addition of 15.00 mL of reagent for the titration of 50.00 mL of 0.0800 M NAOH with 0.1000 M HCI at 25°C. 10 poir QUESTION 7 Calculate the hypothetical pH AFTER addition of 40.00 mL of reagent for the titration of 50.00 mL of 0.0800 M NAOH with 0.1000 M HCI at 25°C.

For acid -base neutralisation

HCl + NaOH ----------------> NaCl + H2O

15x0.1=1.5 50x0.08=4 0 0 initial moles

0 2.5 4 - after addition

The solution has excess base

and [base] = oH-] = mmoles/ volume = 2.5/(15+50) =0.0384

and pOH = -log [OH-] = -log 0.0384 = 1.414

and pH =14-1.414=12.586

Q2)

After 40mL

  

HCl + NaOH ----------------> NaCl + H2O

40x0.1=4 50x0.08=4 0 0 initial moles

0 0 4 - after addition

The solution has no acid no base, only salt of strong acid - strong base which is neutral.

Thus pH = 7.0

3)

After 46mL

HCl + NaOH ----------------> NaCl + H2O

46x0.1=4.6 50x0.08=4 0 0 initial moles

0.6 0 4 - after addition

The solution has excess acid

and [acid] = [H+] = mmoles/ volume = 0.6/(46+50) =0.00625

and pH = -log [H+] = -log 0.00625 = 2.204