Question

1.A 25.00 ml smaple of 0.523 M nitrous acid, HNO2, solution is titrated with a 0.213 M NaOH. For HNO2, Ka = 4.0 X 10^-4

a) What is the pH before any NaOH is added?

b) Write the reaction that takes place as KOH solution is added to the HNO2 solution.

c) write the reaction that determines the pH at the equivialnce point.

-What is the pH at the equivilance point

-what is the pH at the 1/2 equivilance point

d) sketch a titration curve for this titration. Be sure to label you axes and the three points you determined above.

Answer 1

(a) pH before NaOH is added

pH = 1/2 ( pKa - log [acid] )

pKa = -log Ka = -log (4.0 X 10^-4) = 3.39

pH = 1/2 * (3.39 - log (0.523) ) = 1.83

(b) Reaction

NaOH + HNO2 = NaNO2 + H2O

(c) pH at equivalence point

at equivalence point mmols of acid and base are equaland they react completely to form salt

mmol = molarity * volumein mL

mmols of acid = 25 * 0.523 =13.075

mmols of NaOH = 13.075 hence volume of NaOH = 13.075 / 0.213 = 61.38 mL

total volume = 61.38 + 25 = 86.38 mL

[salt] = mmol / total volume = 13.075 / 86.38 = 0.151 M

hence pH of salt of weak acid nd strong base is given by

pH = 1/2 * ( pKw + pKa + log [salt])

pH = 1/2 * ( 14 + 3.39 + log (0.151) ) = 8.28

at mid point

pH = pKa since acid and its conjugate salt concentration are equal

pH = 3.39

Titration curve