Question

Determine the pH at the equivalence point for the titration of 50 mL of HNO2 and KOH with concentration of .1 M and .1 M respectively.

Answer 1

KOH + HNO2 ---> KNO2 + H2O

first find out the concentration of KNO2

no of moles of HNO2 = molarity x volume in liters = 0.1M x 0.05 L = 0.005 moles of HNO2

no of moles of KOH = 0.1M x 0.05L = 0.005 moles

from the balanced iquation it is clear that from noe mole of each one mole ofKNO2 will come

so no o fmoles of NaNO2 = 0.005 moles KNO2

total volume = 0.05L + 0.05L = 0.1L

molarity of KNO2 = no of moles / volume in liters = 0.005 / 0.1 = 0.05 M

KNO2 can dissociate as a

KNO2 + H2O ---> HNO2 + KOH

K hydrolysis = K water / K acid = 10^-14 / 5.1x10^-14 = 1.96 x 10^-11 (Ka of HNO2 taken from google)

1.96 x 10^-11 =  [HNO₂] [OH^-] / [KNO₂^-]

1.96 x 10^-11 = [x][x] / 0.05

x² = 9.8 x 10^-13

x = 9.9 x 10^-7 = [OH-]

pOH = log[9.9 x 10^-7] = 6.00

pH = 14- pOH = 14-6.00 = 8