pH problem...please help?

Question

I have no idea how to start solving this...

I need to find the pH of solution A and C.
You don't need to know the identity of X, except that the X in solution A is the same as in solution B.
All solutions are at 25°C.

Solution A is 50.0 mL of a 0.150 M solution of the weak monoprotic acid HX.
Solution B is a 0.0500 M solution of the salt NaX. It has a pH of 10.65.
Solution C is made by adding 15.0 mL of 0.250 M KOH to solution A.

Please help, thank you! :)

Answer 1

First ask yourself: What do I need to find the pH of solution A?

You know you have a weak monoprotic acid.
You know the initial concentration of the acid C=0.150 M
so you need to find the Ka

How can we find that? From solution B. The salt NaX contains the anion X- which is the conjugate base of the weak acid HX. Thus it will hydrolyze according to the reaction
X- + H2O <=> HX + OH- with Kb=Kw/Ka

So we will use the info we have on solution B to find the Kb and from that we will find the Ka

.. .. .. .. .. .. X- + H2O <=> HX + OH-
Initial .. .. 0.05
React .. .. .. x
Produce .. .. .. .. .. .. .. .. .. .. x .. .. x
At Equil .. 0.05-x.. .. .. .. .. . .x .. .. x

Kb= [HX][OH-] / [X-] = x^2/(0.05-x)

but x=[OH-] and [OH-]=10^-pOH so x=10^-pOH
Also pOH= 14-pH= 14-10.65 = 3.35
thus x=10^-3.35 and
Kb= ((10^-3.35)^2) / (0.05-10^-3.35) = 4.03*10^-6

Kb=Kw/Ka => Ka=Kw/Kb = (10^-14)/(4.03*10^-6) = 2.48*10^-9

Now we can find the pH of A

.. .. .. .. .. .. HX <=> H+ +X-
Initial .. .. ..0.15
Dissoc. .. .. y
Produce .. .. .. .. .. .. y .. ..y
At Equil .. 0.15-y .. ..y .. ..y

Ka= [H+][X-] / [HX] = y^2/(0.15-y)
Let's assume that 0.15 >> y so that 0.15-y=0.15
Then the equation is simplified to

Ka= y^2/0.15 =>
y= squareroot (0.15*Ka) = SQRT(0.15*2.48*10^-9) = 1.93*10^-5 M

pH= -log[H+] =-logy= -log(1.93*10^-5) = 4.71

For C we need to find out the moles of each compound and see what remains after the acid-base reaction

mole HX= M1*V1= 0.15 *50*10^-3 = 7.5* 10^-3
mole KOH = M2*V2 = 0.25*15*10^-3 = 3.75*10^-3

The reaction is
HX +KOH -> KX +H2O

We see that the stoichiometry is 1 HX : 1 KOH : 1 KX.
thus the base will react completely and some acid will remain.
The base is the limiting reagent thus we use it for finding the amounts of the compounds in the solution after the reaction

formed mole X- = mole KX= mole KOH = 3.75*10^-3
remaining mole HX= initial-reacted = 7.5*10^-3 - 3.75*10^-3 = 3.75*10^-3

C is a solution of the weak acid HX and its salt with a strong base, thus it is a buffer. In order to find the pH there are 2 ways. The exact and an approximate.
The exact would be to find the concentrations by dividing with the new volume V=V1+V2= (50+15)*10^-3 = 65*10^-3 L, set up an ICE table for the dissociation of HX with initial concetrations for both HX and X- and solve the resulting quadratic.

The approximate is using the Henderson-Hasselbalch equation (and approximations) for buffers. I'm pretty sure you will reach the same result since Ka is so small and C is quite big. The equation is

pH= pKa+log[conj.base]/[acid] =>
pH= pKa + log [X-]/[HX] = pKa +log[KX][HX] (This is because KX dissociates fully with 1:1 stoichiometry)

the ratio of concentrations is also a ratio of moles since volume is simplified in the ratio (thus we avoid doing some calculations), thus

pH= pKa +log( mole KX / mole HX)
and assuming that the initial concentrations are practically equal to the equilibrium concentrations:
pH = pKa +log3.75*10^-3 /3.75*10^-3 =pKa+log1= pKa =-logKa = -log(2.48*10^-9) = 8.61

Answer 2

Solution B

NaX >> Na+ + X-

X- + H2O <> HX + OH-
At equilibrium

pH=10.65
pOH=3.35>>>>(OH-)=10^-3.35=0.000447 M =(HX)
(X-)=0.0500-0.000447=0.0495

Kh=Kw/Ka= (0.000447)^2 / 0.0495 =4.04 10^-6

Kw/4.04 10^-6 =Ka

Ka=2 10^-9


Solution A

HA <>H+ + A-
initial concentration
0.15.....0......0
at equilibrium
0.15-x...x.......x

Ka= (x)(x)/0.15-x = 2 10^-9

x= 0.0000193 M

pH= -log 0.0000193 = 4.71

Solution C
moles HX = 0.0075
moles OH- =0.00375

Total volume = 65 mL=0.065 L

0.0075 moles HX + 0.00375 moles OH- >>0.00375 moles HX and 0.00375 mole X-

initial concentration HX=OH-= 0.00375/0.065=0.0577 M

HX <>X- +H+

Ka=2 10^-9 =(x)(0.0577+x)/0.0577-x)

x=2 10^-9

pH=8.7