In: Chemistry
Write the reacction in question 14 and answer 14-17 based one on the same equation. (Use the ICE table to solve the problems)
14. A 50.0 mL sample of 0.100 M HNO2 is titrated with 0.200 M KOH. Calculate the initial pH of the solution. (Ka = 4.6x10^-4)
15. CalCulate the pH after the addition of 20 mL of 0.200 mL KOH to the titration from the question 14.
16. Calculate the pH at the equivalence point for the titration from question 14.
17. Calculate the pH after the addition of 35 mL of KOH to the titration from question 14.
15 )
millimoles of HNO2 = 50 x 0.1 = 5
millimoles of KOH = 20 x 0.2 = 4
pKa = -log Ka
pKa = -log (4.6 x 10^-4) = 3.34
HNO2 + KOH ---------------------> KNO2 + H2O
5 4 0 0
1 0 4
pH = pKa + log [K NO2 / HNO2]
pH = 3.34 + log (4 /1)
pH = 3.94
16)
volume of base needed at equivalence point = 50 x 0.1 / 0.2 = 25
at equivalence point only salt remains salt concentration = 5 / (25 + 50) = 0.067 M
pH = 7 + 1/2 [pKa + log C]
pH = 7 + 1/2 [3.34 + log 0.067 ]
pH = 8.08