1. HA(aq) + NaOH(aq) < ----> NaA(aq) +
H2O(l)
2. limiting reagent is NaOH.
3.
total volume = 10+5+50 = 65 ml.
No of moles of HA = 10/1000*1 = 0.01 mole
molarity of HA = 10/65*1 = 0.154 M
No of moles of NaOH= 5/1000*0.5 = 0.0025
mole.
molarity of NaOH = 5/65*0.5 = 0.0385 M
4. HA(aq) <---> H+(aq) + A-(aq)
5.  k = [H+][A-]/[HA]
6. pH = -log[H+]
[H+] = 10^(-Ph) = 10^(-5) = 1*10^(-5) M
             Â
HA(aq) <--->Â Â Â H+(aq)Â Â
+Â Â Â A-(aq)
Initial     Â
0.154Â Â Â Â Â Â Â Â Â Â Â
0Â Â Â Â Â Â Â Â Â Â Â
0
equilibrium 0.154 - 1*10^(-5) 1*10^(-5)Â Â
1*10^(-5)
k = (1*10^(-5)*1*10^(-5)) / (0.154 -
1*10^(-5))
   = 6.5*10^-10
8) K = ka no difference.
