What must be the minimum concentration of dichloroacetic acid in 1.00L buffer solution of a pH 0.91 dichloroacetic acid/sodium dichloroacetate buffer if the pH changes by 0.2 units when 0.057 moles HCl are added?
Dichloroacetic Cl2CHCOOH ka= 5.50E-02 pka= 1.26
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It took awhile but here it is.
For pH = 0.91,you have
0.91 = 1.26 + log (base)/(acid)
B/A = approx 0.447 but you need to go through this and all calculations that follow to get more accurate numbers. What you want to do now is to add 57 millimols H^+ to the base(which I will call ClAc^-) so that the pH will change from 0.91 to 0.71 and you want to know the acid (which I will call HClAc) will be for that to happen.
.........ClAc^- + H^+ --> HClAc
I.........B.......0........A
add...............57...........
C........-57.....-57......+57
E........B-57.....0.......A+57
0.71 = 1.26 + log (B-57)/(A+57)
but B = 0.447A
0.71 = 1.26 + log (0.447A-57)/(A+57)
Solve for A. I get approx 440 millimols which in 1000 mL will be 0.44M and since the base is 0.447 x that it will be approx 197 millimols or 0.197 M. I like to check these things to make sure it works. It may not be exactly 0.71 because I've rounded here and there but if you go through it will more accurate numbers it should come out ok.
.......ClAc^- + H^+ ==> HClAc
I.......197.....0........440
add............57...........
C.......-57...-57.........+57
E.......140....0.........497
pH = 1.26 + log(140/497)
pH = 0.7098 which rounds to 0.71
To answer this question, we need to use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa of the acid and the ratio of the concentration of the acid to its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the desired pH value
pKa = the dissociation constant of the acid
[A-] = concentration of the conjugate base (in this case, the concentration of sodium dichloroacetate)
[HA] = concentration of the acid (in this case, the concentration of dichloroacetic acid)
Given:
pH = 0.91 (initial pH before adding HCl)
pKa = 1.26
[A-] = ? (concentration of sodium dichloroacetate)
[HA] = ? (concentration of dichloroacetic acid)
ΔpH = 0.2 (change in pH when 0.057 moles HCl are added)
First, we need to calculate the initial concentration of dichloroacetic acid ([HA]) and sodium dichloroacetate ([A-]).
1. Calculate the initial concentration of dichloroacetic acid ([HA]):
We can rearrange the Henderson-Hasselbalch equation to solve for [HA]:
[HA] = 10^(pH - pKa) * [A-]
Substituting the values:
[HA] = 10^(0.91 - 1.26) * [A-]
2. Determine the change in concentration of dichloroacetic acid ([HA]) when 0.057 moles of HCl are added:
Using the balanced chemical equation for the reaction between HCl and dichloroacetic acid, we can determine the stoichiometry between the two:
2 HCl + dichloroacetic acid -> 2 sodium dichloroacetate
From the balanced equation, we can see that the stoichiometric ratio between HCl and dichloroacetic acid is 2:1. Therefore, for every mole of HCl added, the concentration of dichloroacetic acid decreases by 2 moles.
So the change in concentration of dichloroacetic acid ([HA]) is:
Δ[HA] = -2 * 0.057
3. Calculate the new concentration of dichloroacetic acid ([HA]):
[HA]final = [HA]initial + Δ[HA]
Now we have the new concentration of dichloroacetic acid ([HA]) in the buffer solution.
4. Use the Henderson-Hasselbalch equation to find the concentration of sodium dichloroacetate ([A-]):
Substituting the values into the Henderson-Hasselbalch equation:
0.91 = 1.26 + log([A-]/[HA]final)
Rearranging the equation:
[A-] = 10^(0.91 - 1.26 + log([HA]final))
Finally, we can substitute the calculated values to find the concentration of sodium dichloroacetate ([A-]) in the buffer solution.