I just would like to know the work done as to how to get those answers for c and
ID: 1029662 • Letter: I
Question
I just would like to know the work done as to how to get those answers for c and d. Thanks Name (Calculators optional - you shouldn't need it) A particular acid, HPar has an equilibrium constant, ID# Ka-1.0 x 10-z. Where z is Kb-lo" ta) s a i the last digit of your ID is 0, 1. or 23 lo1o h9 if the last digit of your ID is 6,7,8 or 9. 5 (1-97,5 7 if the last digit of your ID is 3, 4 or 5 Determine the pH of a solution which is: a) 0.1 M in HPar b) 0.1 M in HPar and 0.1 M in NaPar HepK1 pa.pt. 4 2 c)0.1 M in NaPar d) at the equivalence point when 0.2 M Hpar is titrated with 0.2 M NaOH (note that the total volume will increase)
Explanation / Answer
You need the answers to parts (c) and (d). I do not know you ID; I will assume z = 5 for demonstration purposes.
(c) We have Ka (HPar) = 1.0*10-5.
We know that Ka*Kb = Kw
where Kb = base ionization constant of Par- (conjugate base of weak acid HPar, NaPar is the sodium salt of the conjugate base); Kw = 1.0*10-14 is the ionization constant of water.
Plug in values and obtain
Kb = Kw/Ka = (1.0*10-14)/(1.0*10-5) = 1.0*10-9
Consider the ionization of NaPar (Par- is the active species) as
Par- (aq) + H2O (l) -------> HPar (aq) + OH- (aq)
Since OH- is formed, we work with Kb. We define Kb as
Kb = [HPar][OH-]/[Par-] = (x).(x)/(0.1 – x) (x = concentration of Par- = concentration of H+ due to 1:1 nature of ionization)
====> 1.0*10-9 = x2/(0.1 – x)
Since Kb is small, we can assume (0.1 – x) 0.1 and write
1.0*10-9 = x2/(0.1)
====> x2 = 1.0*10-9*0.1 = 1.0*10-10
====> x = 1.0*10-5
Therefore, [OH-] = 1.0*10-5 and hence, pOH = -log [OH-] = -log (1.0*10-5) = 5.0.
We know that pH + pOH = 14; therefore,
pH = 14 – pOH = 14 – 5 = 9 (ans).
(d) Write down the balanced chemical equation for the reaction as
HPar (aq) + NaOH (aq) --------> NaPar (aq) + H2O (l)
As per the stoichiometric equation,
1 mole HPar = 1 mole NaOH = 1 mole NaPar
We have a 1:1 molar ratio of HPar:NaOH and hence, the volumes required will be the same. The total volume of the system is twice the initial volume.
Let V L of 0.2 M HPar be V L of 0.2 M NaOH. Therefore, we have
moles HPar = moles NaOH = moles NaPar = 0.2*V mole and the total volume is 2V L.
Therefore, concentration of NaPar at equilibrium = (0.2*V mole)/(2V L) = 0.1 M. The problem now becomes the same as in part (c) above and hence, the pH of the solution is 9 (ans).