Question: A 0.400 g sample of propionic acid was dissolved in water to give 50.00 mL of solution. This solution was titrated with 0.150 M NaOH. what was the pH of the solution when the equivalence point was reached?
I'm not really sure how i would go about setting this up. Would i use an ice table? What do i do with the 0.400 g
0.400 g/molar mass propionic acid gives you the moles of propionic acid that you started with. Now that is titrated with 0.150 M NaOH. How much NaOH will it take. Moles NaOH will be same as the moles of propionic acid (since they react in 1:1 ratio) so moles NaOH you know and that = M x L. Knowing moles and M you calculate L NaOH. So the salt produced from the titration: Letting HP stand for propionic acid:
HP + NaOH ==> NaP + HOH.
You can do the ice table if you wish to determine how much NaP is present at the equivalence point but I can tell you before you do it that moles NaP will be the same as the moles HP you started with. So what is the concentration of the NaP? That will be moles NaP/volume at that point (take the volume of NaOH and add to 50 mL--the volume initially used.) The pH at the equivalence point is determined by the hydrolysis of the salt (in this case the NaP). So write a hydrolysis equation, set up an ICE table, and solve for OH^-.
P^- + HOH ==> HP + OH^-
ICE table here.
Kb for P^- is Kw/Ka for propioinic acid.
To find the pH at the equivalence point, we first need to calculate the moles of propionic acid in the solution.
Step 1: Calculate the moles of propionic acid
The molar mass of propionic acid is 74.08 g/mol.
Moles = mass/molar mass = 0.400 g / 74.08 g/mol = 0.00540 mol
Step 2: Calculate the initial concentration of propionic acid
The volume of the solution is 50.00 mL, which is equivalent to 0.0500 L.
Initial concentration = moles/volume = 0.00540 mol / 0.0500 L = 0.108 M
Step 3: Determine the reaction between propionic acid and NaOH
Propionic acid (CH3CH2COOH) reacts with NaOH (sodium hydroxide) to form sodium propionate (CH3CH2COONa) and water (H2O).
CH3CH2COOH + NaOH → CH3CH2COONa + H2O
Step 4: Calculate the volume of 0.150 M NaOH required for neutralization
Since NaOH is a strong base, it will neutralize the acid completely. So, the moles of NaOH required for neutralization are equal to the moles of propionic acid.
Volume of NaOH = moles/initial concentration of NaOH = 0.00540 mol / 0.150 M = 0.0360 L = 36.0 mL
Step 5: Determine the volume of NaOH required to reach the equivalence point
The volume of NaOH required to reach the equivalence point is equal to the volume of propionic acid solution, which is 50.00 mL.
Step 6: Calculate the total volume of the solution at the equivalence point
Total volume = volume of propionic acid solution + volume of NaOH = 50.00 mL + 50.00 mL = 100.0 mL = 0.100 L
Step 7: Calculate the concentration of sodium propionate at the equivalence point
At the equivalence point, all the propionic acid is neutralized, so the concentration of sodium propionate will be:
Concentration = moles/volume = 0.00540 mol / 0.100 L = 0.0540 M
Step 8: Calculate the pOH at the equivalence point
pOH = -log10([OH-])
Since sodium propionate is the salt of a weak acid and a strong base, it completely dissociates in water, providing OH- ions in the solution.
[OH-] = concentration of OH- = concentration of sodium propionate = 0.0540 M
pOH = -log10(0.0540) = 1.267
Step 9: Calculate the pH at the equivalence point
pH + pOH = 14
pH = 14 - pOH = 14 - 1.267 = 12.733
So, the pH of the solution at the equivalence point is approximately 12.733.
To determine the pH of the solution at the equivalence point, we need to understand the reaction that occurs between propionic acid (C2H5COOH) and NaOH.
First, let's convert the mass of the propionic acid into moles. The molar mass of propionic acid is 74.08 g/mol.
Mass of propionic acid = 0.400 g
Moles of propionic acid = (0.400 g) / (74.08 g/mol)
Now, let's consider the balanced chemical equation between propionic acid and NaOH.
C2H5COOH + NaOH → C2H5COONa + H2O
From the balanced equation, we can see that one mole of propionic acid reacts with one mole of NaOH, producing one mole of sodium propionate and one mole of water.
Since the moles of propionic acid and NaOH are equal at the equivalence point, we can use the stoichiometry to find the volume of NaOH solution required to reach the equivalence point.
Moles of NaOH = Moles of propionic acid = (0.400 g) / (74.08 g/mol)
Now, let's determine the volume of NaOH solution using its molarity.
Molarity of NaOH (M1) = 0.150 M
Volume of NaOH solution (V1) = (Moles of NaOH) / (Molarity of NaOH)
Now, we have the volume of NaOH solution required to reach the equivalence point. However, we need to find the pH at the equivalence point.
At the equivalence point, the moles of propionic acid are completely neutralized by moles of NaOH. Hence, there is no excess propionic acid or NaOH present in the solution. Therefore, we can assume that the resulting solution contains only water and the sodium propionate salt.
Sodium propionate is the sodium salt of propionic acid, which is a weak base. When it dissolves in water, it undergoes hydrolysis and produces hydroxide ions (OH-). The hydroxide ions will increase the pH of the solution.
To calculate the pH at the equivalence point, you need to determine the concentration of the hydroxide ions produced by sodium propionate. This can be done by using the balanced chemical equation and considering the stoichiometry.
From the balanced equation: 1 mole of propionic acid reacts with 1 mole of NaOH, producing 1 mole of sodium propionate.
Since the moles of propionic acid and NaOH are equal, the moles of sodium propionate formed will also be equal to the moles of propionic acid and NaOH.
Now, calculate the concentration of hydroxide ions (OH-) using the volume of the NaOH solution and the moles of sodium propionate.
Concentration of hydroxide ions (OH-) = (Moles of sodium propionate) / (Volume of NaOH solution)
Finally, use the concentration of hydroxide ions (OH-) to calculate the pOH and then convert it to pH.
pOH = -log10(Concentration of hydroxide ions)
pH = 14 - pOH
By following these steps, you can determine the pH of the solution when the equivalence point is reached.