Question

# Determine the pH at the equivalence point for the titration of 50 mL of HNO2 and...

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.

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