Question: v cm3 of a 0.2 M HX solution and v cm3 of 0.1 M KOH solution is mixed.pH of the final solution is 5.3

Question

1)Find ka of HX
2)Find the moles of NaOH and HCl should be added respectively,for the pH of 1dm3 of the solution to increase and decrease by 1

My thoughts on the question:

Reactions:(1) HX+KOH-->KX + H2O
(2)HX<===>H^+ + X^-
(3) KX-->K^+ + X^-
(4)X^-<===> HX + OH^-

As the pH of the final solution is 5.3 I assumed that all KOH reacts(is that correct?)
So from (1) the moles of remaining HX=KX formed = [(0.2)-(0.1)]*v*(10)^-3 = v*(10)^-4

(2) HX<==> H+ + X-
equ. (0.1-a) a [a+( (v)(10)^-4)]

(3) KX ----> K+ + X-
equ. v*(10)^-4 v*(10)^-4 [a+(v*(10)^-4)]

For the first part, as this is a buffer solution, applying Henderson equation I got pKa=5.3 and ka=(5.012)*(10)^-6

For the second part I took the moles of NaOH should be added as x.
Then I noted that as this question doesn't tells us about a change of temperature, the ka value should remain the same.So I realized,from Henderson equation, for pH to be increased by 1,the fraction {[lg(acid)]/[lg(salt)]} should be decreased by 1.

(lg=log10)

And for the part about about adding HCl I realized that the same method can be applied.

But when we are equating that fraction to (-1),(as I considered it is being decreased by 1),should we consider both acid's and salt's reactions with the NaOH added?And also we get two unknowns

For example:

As x is the moles of NaOH added,

NaOH+ KX---> KOH + H2O
(x-x) ((v*(10)^-4)-a-x) (+x) (+x)
I think we can neglect "a" and continue calculations.

NaOH + HX ---> NaX + H2O
(x-x) (0.1-a-x) (+x) (+x)

And also we can't simply say all x moles of NaOH reacts with both of them(then we should add 2x moles of NaOH to react with both HX and KX)

What's concentration should be applied as salt's concentration in the Henderson equation? HX or KX??

I'm having too many questions with this problem. Hope someone will explain...

Answer 1

Yes, KOH is used up. Since the volume added is the same we can keep the molarity.
.....HX + KOH ==> KX + H2O
I....0.2...0.1.....0....0
C...-0.1..-0.1...+0.1....0
E....0.1....0......0.1..0.1

pH = pKa + log base/acid
5.3 = pKa + log 0.1/0.1
so pKa = 5.3

For part 2 to change pH by +/- 1.
If you have 1 dm^3, then you have 100 mmols HX and 100 mmols KX.
.....HX + KOH ==> KX + H2O
I....100...0.....100
add........x...........
C.....-x..-x......+x
E...100-x...0......100+x

6.3 = 5.3 + log (100+x)/(100-x) and solve for x = mmols KOH to add.

For the addition of HCl it is
......KX + HCl ==> HX + KCl
I.....100...0......100
add.........x..........
C.....-x...-x......+x
E....100-x..0.....100+x

Substitute into the HH equation and solve for x = mmols HCl added.

I estimated that x should be about 82 millimols NaOH or HCl that should be added to change the pH from 5.3 to either 6.3 or 4.3 but I didn't check my math. You should confirm that.

Answer 2

But we are adding NaOH,not KOH.So can we simply apply the moles added(x) into,
HX+KOH--->KX+H2O ??

Or we can simply do that because both are providing OH- ions to the solution?

Answer 3

Sorry I goofed but all the numbers stay the same. I should have used NaOH and not KOH but both are strong bases and neither Na ion nor K ion are involved. Both are spectator ions anyway. Usually I don't use the cations anyway and write HX + OH^- ==> X^- + H2O or
X^- + H^+ ==> HX. It's the net ionic equation that matters.

Answer 4

Thank you very much! Now I see that I've made too many bad mistake while solving this.

1) considering the hydrolysis of KX(K+ is a strong base action)

2) considering a reaction of a acid with another acid(HX+HCl)

3)considering a reaction of a base with another base(KOH+NaOH)