What volume of 0.5 M NaOH is needed to perform the titration of 30 mL of 0.1 M H

Question

What volume of 0.5 M NaOH is needed to perform the titration of 30 mL of 0.1 M H3PO4? Question options: a) V = 12 mL b) V = 6 mL c) V = 18 mL d) V = 30 mL he pH at the equivalence point when a 0.20 M weak base (Ka = 9.1 x 10-7) is titrated with a 0.20 M strong acid is: Question options: a) pH = 2.9 b) pH = 1.7 c) pH = 4.1 d) pH = 3.5 A student titrates a 25 mL of an unknown concentration of HCl with 35 mL of a 0.890 M solution of KOH to reach the equivalence point. What is the pH of the HCl solution? Question options: a) -0.10 b) 0.10 c) 1.2 A student conducts a titration of 50 mL of HCl with 1.00 M NaOH. The pH at the equivalence point is: Question options: a) Basic b) We can't tell without knowing the volume of NaOH used to reach the equivalence. c) Acid d) Neutral A student discovers a bottle of a highly corrosive monoprotic acid with the label partially destroyed. How can the concentration of the unknown acid be determined without using a pH meter? Question options: a) The student can set-up a titration experiment, where a standardized NaOH solution is used to titrate the unknown strong acid. The moles of OH- added would be equal to moles of H+ in the unknown at the equivalence point, therefore, the concentration of the unknown can be calculated. b) The student can set-up a titration experiment, where a standardized NaOH solution is used to titrate the unknown strong acid. The pH at the equivalence point should be 7 since this is a neutralization reaction, therefore, the concentration of the unknown acid is 1 x 10-7 M. c) The student can set-up a titration experiment, where a standardized NaOH solution is used to titrate the unknown strong acid. The moles of OH- added would be equal to moles of H+ in the unknown at the half-way point, therefore, the concentration of the unknown can be calculated.

Explanation / Answer

1)

moles of H3PO4 = 30 x 0.1/1000 = 3 x10^-3

3 NaOH + H3PO4   ---------------> Na3PO4 + 3 H2O

from this balanced equation :

1 mol H3PO4 -------------->   3 mol NaOH

3x10^-3 mol H3PO4 ----------------> 9 x10^-3 mol NaOH

moles of NaOH = molarity x volume

9 x10^-3 = 0.5 x V

V = 0.018 L

volume of NaOH = 18 mL

option c) V = 18 mL

2)

pKa = -log Ka = -log (9.1 x 10^-7)

       = 6.04

at equivalence point salt only reamins. this salt is formed from weak base and strong acid . so pH < 7

pH = 7 - 1/2 (pKb + log C)

     = 7 - 1/2 (6.04 + log 0.2)

    = 4.33

option c) pH = 4.1

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