Suppose 50.0 mL of 0.300 M HCl is titrated with 0.500 M NaOH. (Remember this mea

Question

Suppose 50.0 mL of 0.300 M HCl is titrated with 0.500 M NaOH. (Remember this means that the 50.0 mL of 0.300 M HCl would be in the flask initially and the buret would contain the 0.500 M NaOH.) a. Calculate the pH in the flask after 20.0 mL of the 0.500 M NaOH is added from the buret. Show your work. b. Calculate the volume of 0.500 M NaOH needed to reach the equivalence point. c. What would be the pH at the equivalence point of this titration? Explain why you would expect the pH to be this particular value. Suppose 25.0 mL of 0.100 M benzoic acid (a weak acid which can be abbreviated HB_2) is titrated by 0.200 M NaOH. (K_a of benzoic acid = 6.3 times 10^5) The balanced equation representing the neutralization would be: HBz (aq) + NaOH (aq) rightarrow NaB_2 (aq) + H_2O (I) a. Calculate the pH after addition of 4.0 mL of 0.200 M NaOH. Show your work. b. Calculate the pH at the equivalence point of this titration. Show your work. c. Calculate the pH when 5.00 mL of 0.20

Explanation / Answer

a) We need to determine the moles of H+ remaining and the volume of the solution to calculate [H+].

moles = [concentration] x volume

moles OH- = (0.5 mol/L) x (0.020 L) = 0.01 mol OH-

starting moles H+ = (0.3 mol/L) x (0.05 L) = 0.015 mol H+

moles H+ remaining after addition of 10 mL base = 0.015 – 0.010 = 0.005 mol H+

volume after addition of 20.0 mL base = 50.0 mL acid + 20.0 mL base = 70 mL = 0.07 L

[H+] = 0.005 mol / 0.07 L = 7.14 x 10-2 M H+

pH = -log (7.14 x 10-2) = 1.15

b) moles OH- = 0.015 (same as no. Of moles of HCl)

So, moles OH- = CONC * VOLUME
0.015 moles / 0.5 mol L-1 = volume

0.03 L = volume

Or, 30 mL = volume

C) moles H+ remaining = 0.015 – 0.015 = 0

[H+] = [OH-] pH = 7.00

There is a large change in the pH at the equivalence point. The pH jumps from about 2 to 7.

pH will be 7 because it is a strong acid and strong base titration.

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