How many milliliters of 0.202 M KOH should be added to 500.0 mL of 0.0233 M
tartaric acid (C4H6O6; FW 150.087) to adjust the pH to 2.75? Ka1 = 1.0 x 10-13 and Ka2 = 4.6 x10-5.
For this one, the Ka1 seems sort of off. The pka would be 13 which doesn't make sense.
Tartaric acid is a dibasic acid (two carboxyl COOH groups). You want to add enough acid to completely neutralize the first H and leave the second. That formula is too long to write; let's call it H2C. We add KOH to neutralize the first H to make KHC. Then we make the buffer out of the KHC and C^2-
mL x M = mL x M
500 x 0.0233 = mL x 0.202 which gives you about 11.65 (but you can do it more accurately and watch the significant figures). Then we make the buffer from that.
........HC^- + OH^- ==> C^2- + H2O
I.......11.65...0.......0........0
add.............x.................
C........-x....-x.......x.........x
E.......11.65-x..0.......x........x
Plug all of this into the HH equation and solve for x.
2.75 = pK2 + log(base)/(acid)
base = x
acid = 11.65-x
x = mmols KOH. Convert to mL of the 0.202M stuff and ADD to the amount needed for the first equivalence point.
Then I would work backwards, plug in the numbers, and see if it really does produce a pH of 2.75.
I ended up with KOH mL being about 1.5mL. The equiv. point would be around 57mL. I see the answer is to be around 20mL. I have no idea how this comes about.
To calculate the amount of KOH needed to adjust the pH, we need to consider the dissociation of the tartaric acid (C4H6O6) and the subsequent reaction with KOH. However, before proceeding with the calculation, let's address your concern about the pKa value of Ka1.
The pKa value of Ka1 can indeed seem off as it is very high. However, it is important to note that Ka values are usually expressed in scientific notation, and in this case, it is 1.0 x 10^(-13). The negative exponent indicates that this is an extremely weak acid with a small ionization constant. The pKa is calculated by taking the negative logarithm of the Ka value, so the pKa for Ka1 in this case would be 13. Please note the negative sign when converting from Ka to pKa.
Now, let's proceed to calculate the amount of KOH needed to adjust the pH to 2.75 using the given information.
First, we need to determine the equilibrium concentrations of [C4H6O6] and [C4H5O6^-]. To do this, we will need the Kb value since tartaric acid is a weak acid.
Kb = Kw/Ka (where Kw is the ion product of water)
Given that Kw = 1.0 x 10^(-14), we can calculate Kb for tartaric acid:
Kb1 = Kw/Ka1 = (1.0 x 10^(-14))/(1.0 x 10^(-13)) = 0.1
Next, we need to determine the concentrations of [C4H6O6] and [C4H5O6^-] at equilibrium using the Henderson-Hasselbalch equation:
pOH = pKb + log([C4H5O6^-]/[C4H6O6])
Since pOH + pH = 14, we can calculate the pOH:
pOH = 14 - pH = 14 - 2.75 = 11.25
Now, we can rearrange the Henderson-Hasselbalch equation to solve for the ratio [C4H5O6^-]/[C4H6O6]:
[C4H5O6^-]/[C4H6O6] = 10^(pOH - pKb)
[C4H5O6^-]/[C4H6O6] = 10^(11.25 - (-log10(Kb1)))
[C4H5O6^-]/[C4H6O6] = 10^(11.25 + 13)
[C4H5O6^-]/[C4H6O6] = 10^(24.25)
[C4H5O6^-]/[C4H6O6] ≈ 1.78 x 10^(24.25)
Now, we can determine the total concentration of tartaric acid (C4H6O6) initially present:
Total concentration of C4H6O6 = moles of C4H6O6 / volume of solution
Given the volume is 500.0 mL and the molarity is 0.0233 M, we can calculate the moles of C4H6O6:
moles of C4H6O6 = molarity x volume
moles of C4H6O6 = 0.0233 mol/L x 0.500 L = 0.01165 mol
Therefore, the total concentration of C4H6O6 is 0.01165 moles / 0.500 L = 0.0233 M
To determine the amount of KOH needed, we need to neutralize the acidic form [C4H5O6^-] of tartaric acid with a base, KOH. The neutralization reaction is as follows:
H^+ (from KOH) + C4H5O6^- → H2O + C4H6O6
From the balanced reaction, it is evident that the molar ratio is 1:1 between KOH and C4H5O6^-.
Therefore, the number of moles of KOH required is also 0.01165 moles.
Now we can calculate the volume of KOH solution needed:
Molarity of KOH = moles of KOH / volume of KOH solution
Given that the molarity of KOH is 0.202 M, we can rearrange the equation to calculate the volume:
volume of KOH solution = moles of KOH / Molarity of KOH
volume of KOH solution = 0.01165 mol / 0.202 mol/L ≈ 0.0576 L
Finally, converting the volume to milliliters:
volume of KOH solution = 0.0576 L x 1000 mL/L ≈ 57.6 mL
Therefore, approximately 57.6 milliliters of 0.202 M KOH should be added to 500.0 mL of 0.0233 M tartaric acid to adjust the pH to 2.75.