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
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...
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.
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?
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.
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)
To find the Ka of HX, we can use the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]).
From the information given, we know that the pH of the final solution is 5.3. We can assume that all of the KOH reacts with HX, as you mentioned. Thus, all of the HX is converted to KX.
Using the Henderson-Hasselbalch equation for the equilibrium reaction HX ⇌ H+ + X-, we can set up the equation:
5.3 = pKa + log([X-]/[HX])
Since we know that all of the HX has been converted to KX, we can substitute [X-] with the concentration of KX. Let's call this concentration [KX].
5.3 = pKa + log([KX]/[HX])
Now, let's consider the moles of HX and KX. We know that the initial concentration of HX is 0.2 M and the final volume of the solution is v cm^3 (or v * 10^-3 L). So, the initial number of moles of HX is 0.2 * (v * 10^-3) mol.
Since all of HX has been converted to KX, the final number of moles of KX is also 0.2 * (v * 10^-3) mol.
Using the equation, we can write:
5.3 = pKa + log([KX]/(0.2 * (v * 10^-3)))
Solving for pKa:
pKa = 5.3 - log([KX]/(0.2 * (v * 10^-3)))
Therefore, the Ka of HX can be calculated as 10^(-pKa).
For the second part of the question, if you want to increase the pH of the solution by 1 unit, you can use the Henderson-Hasselbalch equation again:
pH = pKa + log([A-]/[HA])
In this case, you want to increase the pH by 1, so the new pH would be 6.3. Let's call the new concentration of A- as [A-]' and the new concentration of HA as [HA]'. Now we can set up the equation:
6.3 = pKa + log(([A-]')/([HA]'))
Since we are looking for the change in concentration, we can subtract the old concentration from the new concentration:
6.3 = pKa + log(([A-]' - [KX])/(0.2 * (v * 10^-3) - [HX]))
Similarly, if you want to decrease the pH by 1 unit, you can use the same equation with the new pH value of 4.3.
Please note that the calculations may get more complex if you also consider the reactions with NaOH and HCl. It might be necessary to consider the reactions separately and calculate the moles of NaOH and HCl added based on those reactions. Additionally, the concentration to be used in the Henderson-Hasselbalch equation should be [HX] or [KX], depending on the specific reaction you are considering.
I hope this explanation helps! Let me know if you have any further questions.