Calculate the pH during the titration of 30.00 mL of 0.1000 M NaOH(aq) with 0.1000 M HCl(aq) after 29.3 mL of the acid have been added.

Question

Calculate the pH during the titration of 30.00 mL of 0.1000 M NaOH(aq) with 0.1000 M HCl(aq) after 29.3 mL of the acid have been added.

the answer is 11.07

please explain your steps, thank you :)

Answer 1

HCl + NaOH ====> NaCl + H2O

moles NaOH = 0.030000 L x 0.1000 mol/L = 0.003000 moles

moles HCl = 0.0293 L x 0.1000 mol/L = 0.00293 moles

excess moles of NaOH = 0.003000 - 0.00293 = 0.00007 moles NaOH

final volume = 30.00 ml + 29.3 ml = 59.3 ml = 0.0593 L

Final [NaOH] = 0.00007 moles/0.0593 L = 0.001180 M NaOH = 0.001180 M OH^-

pOH = -log [OH^-] = -log 0.001180 = 2.928

pH = 14 - pOH = 14 - 2.928 = 11.07