HCl is a strong acid and NaOH is a strong base. At what pH do you expect the equ

Question

HCl is a strong acid and NaOH is a strong base. At what pH do you expect the equivalence point of the titration to occur? Around what pH value should the indicator used in this experiment change color?

Explanation / Answer

magine an experiment in which 0.10M HCl is added 1mL at a time to a conical flask containing 10mL 0.10M NaOH solution. HCl(aq) + NaOH(aq) -----> NaCl(aq) + H2O(l) Calculate the pH of the NaOH(aq) before any HCl is added. [OH-] = [NaOH] = 0.10 mol L-1 pOH = -log10[OH-] = -log10[0.10] = 1 pH = 14 - pOH = 14 - 1 = 13 Calculate the pH of the solution after 1mL 0.10 HCl has been added. (NaOH is in excess, HCl is the limiting reagent) Calculate moles of HCl added: n(HCl) = M x V M = 0.10M V = 1mL = 1 x 10-3L n(HCl) = 0.10 x 1 x 10-3 = 1 x 10-4 mol Calculate moles NaOH unreacted = initial moles NaOH - moles NaOH reacted initial moles NaOH = M x V M = 0.10M V = 10mL = 10 x 10-3L initial moles NaOH = 0.10 x 10 x 10-3 = 1 x 10-3mol moles NaOH reacted = moles HCl added = 1 x 10-4mol moles NaOH unreacted = 1 x 10-3 - 1 x 10-4 = 9 x 10-4mol Calculate [OH-] = n(unreacted OH-) ÷ total volume n(unreacted OH-) = n(unreacted NaOH) = 9 x 10-4mol total volume = 10mL + 1mL = 11mL = 11 x 10-3L [OH-] = 9 x 10-4 ÷ 11 x 10-3 = 0.082 mol L-1 Calculate pOH: pOH = -log10[OH-] = -log10[0.082] = 1.09 Calculate pH: pH = 14 - pOH = 14 - 1.09 = 12.91 Continue these calculations until 11mL 0.10 HCl is added. At this point the NaOH is no longer in excess, rather it is now the HCl that is in excess. Calculate the pH of the solution after 11mL HCl has been added moles HCl: n(HCl) = M x V M = 0.10 mol L-1 V = 11mL = 11 x 10-3L n(HCl) = 0.10 x 11 x 10-3 = 1.1 x 10-3mol Calculate moles HCl in excess n(HCl) unreacted = total n(HCl) - n(HCl) reacted total n(HCl) = 1.1 x 10-3 mol n(HCl) reacted = n(NaOH) = 1 x 10-3 mol n(HCl) unreacted = 1.1 x 10-3 - 1 x 10-3 = 1 x 10-4 mol Calculate [H+]: [H+] = n(H+ unreacted) ÷ total volume n(H+) unreacted = n(HCl) unreacted = 1 x 10-4 mol total volume = 10mL + 11mL = 21mL = 21 x 10-3L [H+] = 1 x 10-4 ÷ 21 x 10-3 =4.76 x 10-3 mol L-1 Calculate pH of the solution pH = -log10[H+] = -log10[4.76 x 10-3] = 2.32 Continue these calculations until all the HCl has been added

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