Strong base is dissolved in 565 mL of 0.400 M weak acid (K, = 4.43 x 105) to mak
ID: 692837 • Letter: S
Question
Strong base is dissolved in 565 mL of 0.400 M weak acid (K, = 4.43 x 105) to make a buffer with a pH of 3.98. Assume that the volume remains constant when the base is added. HA(aq)+OH-(aq) H2O(l) + A-(aq) Calculate the pKa value of the acid and determine the number of moles of acid initially present. Number Number pK, = | | 4.35 226 mol HA When the reaction is complete, what is the concentration ratio of conjugate base to acid? Number How many moles of strong base were initially added? Number mol OH G Previous Give Up & View Solution Check Answer Next Exi_Explanation / Answer
o of moles of HA = molarity * volume in L
= 0.4*0.565 = 0.226 moles
let no of moles of base (OH^-] = x
PKa = -logKa
= -log4.43*10^-5
= 4.3535
HA(aq) + OH^- (aq) -----------> H2O(l) + A^- (aq)
I 0.226 x 0
C -x -x +x
E 0.226-x 0 +x
PH = Pka + log[A^-]/[HA]
3.98 = 4.3535 + log [A^-]/[HA]
log[A^-]/[HA] = 3.98-4.3535
log[A^-]/[HA] = -0.3735
[A^-]/[HA] = 10^-0.3735 = 0.4231
[A^-]/[HA] = 0.4231
x/0.226-x = 0.4231
x = 0.4231*(0.226-x)
x = 0.06719
no of moles of OH^- = x = 0.06719moles