Consider the titration of 100.0mL of 0.10 M malonic acid with 0.10 M NaOH. What is the pH after the addition of 175 mL NaOH?
Let's call malonic acid H2M.
millimols H2M = mL x M = 100 x 0.1 = 10
millimols NaOH = 175 x 0,1 = 175 so you have two equivalence points. The first H is titrated by the first 100 mL of NaOH like this.
.........H2M + OH^- ==> HM^- + H2O
I........10..........0...............0............
add................10................................
C.....-10.........-10.............+10
E.........0............0..............+10
At the end of the first 100 mL you have left 10 mmols HM^- and the next 75 mL titrates the second H like this.
............HM^- + OH^- ==> M^2- + H2O
I...........10..........0..............0..................
add..................7.5....................................
C.........-7.5.....-7.5...........+7.5................
E............2.5......0...............7.5
So this gives you a buffered solution with HM^- as the acid and M^2- as the base. Substitute this into the Henderson-Hasselbalch equation and solve for pH.
pH = pKa2 + log [(M^2-)/(HM^-)]
Post your work if you get stuck.
To determine the pH after the addition of 175 mL of NaOH to the titration of 100.0 mL of 0.10 M malonic acid, we need to find the number of moles of NaOH that reacted with the malonic acid.
Step 1: Calculate the number of moles of malonic acid:
moles of malonic acid = volume (L) × concentration (mol/L)
moles of malonic acid = 100.0 mL × (1 L / 1000 mL) × 0.10 mol/L
moles of malonic acid = 0.010 mol
Step 2: Determine the limiting reagent:
Since the stoichiometry between malonic acid (C3H4O4) and NaOH is 1:1, we can infer that 0.010 mol of NaOH will react with 0.010 mol of malonic acid.
Step 3: Calculate the number of remaining moles of NaOH:
moles of NaOH remaining = total moles of NaOH - moles of NaOH reacted
moles of NaOH remaining = (volume of NaOH added (L) × concentration of NaOH (mol/L)) - moles of malonic acid reacted
moles of NaOH remaining = 0.175 L × 0.10 mol/L - 0.010 mol
moles of NaOH remaining = 0.0175 mol - 0.010 mol
moles of NaOH remaining = 0.0075 mol
Step 4: Calculate the concentration of NaOH remaining:
concentration of NaOH remaining = moles of NaOH remaining / volume of NaOH remaining (L)
concentration of NaOH remaining = 0.0075 mol / 0.175 L
concentration of NaOH remaining ≈ 0.043 M
Step 5: Convert the concentration of NaOH remaining to pOH:
pOH = -log10(concentration of OH-)
pOH = -log10(0.043)
pOH ≈ 1.37
Step 6: Calculate the pH:
pH = 14 - pOH
pH = 14 - 1.37
pH ≈ 12.63
Therefore, the pH after the addition of 175 mL of NaOH is approximately 12.63.
To determine the pH after the addition of 175 mL NaOH, we need to consider the reaction that occurs between malonic acid (HA) and sodium hydroxide (NaOH).
The balanced chemical equation for the reaction is as follows:
HA + NaOH → NaA + H2O
where HA represents malonic acid and NaA represents the sodium salt of malonic acid.
Since malonic acid is a weak acid, it will undergo a partial ionization in water:
HA ⇌ H+ + A-
where H+ is the hydronium ion and A- is the malonate ion.
Initially, we have 100.0 mL of a 0.10 M solution of malonic acid, which means we have (0.10 mol/L) × (0.100 L) = 0.010 mol of malonic acid.
The first step is to determine how many moles of NaOH have been added. To do this, we need to calculate the number of moles of NaOH in 175 mL of the 0.10 M NaOH solution:
(0.10 mol/L) × (0.175 L) = 0.0175 mol NaOH
Since the stoichiometry of the reaction is 1:1 between HA (malonic acid) and NaOH, the number of moles of HA that have reacted can be calculated as 0.0175 mol.
The remaining amount of HA after the reaction is:
0.010 mol - 0.0175 mol = -0.0075 mol
Now we need to calculate the concentration of HA after the reaction. We know that the total volume of the solution is 100.0 mL + 175 mL = 275 mL = 0.275 L. Therefore, the concentration of HA is:
(-0.0075 mol) / (0.275 L) = -0.027 mol/L
Since the concentration of HA is negative, it indicates that the pH is above the range of pH measurements (pH > 14). This means that the pH after the addition of 175 mL of NaOH is greater than 14.
Therefore, the pH cannot be determined directly from the given information, as it exceeds the pH range of the usual pH scale.