In: Chemistry
An aqueous solution contains 0.341 M nitrous acid. How many mL of 0.393 M potassium hydroxide would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 3.140?
no of moles of HNO2 = molarity * volume in L
=0.341*0.15 = 0.05115moles
no of moles of KOH = molarity * volume in L
= 0.393*v
HNO2(aq) + KOH(aq) ---------> KNO2(aq) + H2O(l)
I 0.05115 0.393*v 0
C -0.393*v -0.393*v 0.393*v
E 0.05115-0.393*v 0 0.393*v
PH = 3.14
PKa = 3.15
PH = PKa + log[KNO2]/[HNO2]
3.14 = 3.15 + log0.393*v/( 0.05115-0.393*v)
log0.393*v/( 0.05115-0.393*v) = 3.14-3.15
log0.393*v/( 0.05115-0.393*v) = -0.01
0.393*v/( 0.05115-0.393*v) = 10^-0.01
0.393*v/( 0.05115-0.393*v) = 0.977
0.393*v = 0.977*( 0.05115-0.393*v)
v = 0.0643
volume of KOH = 0.0643L = 64.3ml >>>>answer