Homework SourseA 350 mL sample of 0142 M HNO2 is titrated with 0206 M KOH K
A 35.0 mL sample of 0.142 M HNO2 is titrated with 0.206 M KOH. (Ka for HNO2 is 4.57×104.)
A 35.0 mL sample of 0.142 M HNO2 is titrated with 0.206 M KOH. (Ka for HNO2 is 4.57×104.)
A 35.0 mL sample of 0.142 M HNO2 is titrated with 0.206 M KOH. (Ka for HNO2 is 4.57×104.)
Solution
HNO2 = 35.0ml of 0.142M
number of moles of HNO2= 0.142Mx0.035L= 0.00497 moles
KOH= 0.206M
volume of KOH = 0.142x35.0/0.206 = 24,13ml
number of moles of KOH= 0.206Mx0.02413L = 0.00497 moles
Ka = 4.57x10^-4
-log(Ka) = -log(4.57x10^-4)
PKa= 3.34
It is the equivalent point
so at equivalent point
PH= 7 + 1/2[PKa + logC]
C= number of moles/volume
total volume= 35.0 + 24.13 = 59.13 ml = 0.05913L
C= 0.00497/0.05913 = 0.084M
PH= 7 + 1/2[ 3.34+ log(0.084)]
PH= 8.13
