Question

1. Calculate the pH of a 0.339 M H2S. Calculate the [S2-] in the solution.

2. A titration is performed by adding 0.656 M KOH to 40 mL of 0.172 M HNO₃:

b) Calculate the pH after the addition of 2.1, 5.25 and 9.49 mL of the base.(Show your work in detail for one of the volumes.)

e)Calculate the pH after adding 5.00 mL of KOH past the equivalence point

Answer 1

1 ) for H2A acid [ A-2 ] = Ka2

[ S-2 ] = 1.0 x 10^-19 M

2)

millimoles of HNO3 = 40 x 0.172 = 6.88

after adding 2.1 mL KOH :

millimoles of KOH = 0.656 x 2.1 = 1.38

acid concnetration = (6.88 -1.38) / (40 + 2.1) = 0.131 M

pH = -log [H+] = -log (0.131) = 0.88

pH = 0.88

5.25 mL added

millimoles of KOH = 0.656 x 5.25 = 3.44

acid concnetration = (6.88 -3.44) / (40 + 5.25 ) = 0.076 M

pH = -log [H+] = -log (0.076) = 1.12

pH = 1.12

adding 9.49 mL

millimoles of KOH = 0.656 x 9.49 = 6.23

acid concnetration = (6.88 -6.23) / (40 + 9.49) = 0.0132 M

pH = -log [H+] = -log (0.0132) = 1.88

pH = 1.88

e) equivalece point volume

0.656 x V = 40 x 0.172

V = 10.49 mL

now 5.00 mL of KOH past the equivalence point

now volume = 10.49 + 5 = 15.49 mL

millimoles of KOH = 0.656 x 15.49 = 10.16

base concnetration = (10.16 - 6.88 ) / (40 + 15.49) = 0.0591M

pOH = -log [OH-] = -log (0.0591) = 1.23

pOH = 1.23

pH + pOH = 14

pH = 12.77