What mass of propanoic acid, CH3CH2COOH, pKa= 4.87, should be dissolved in water to 2.00L volume in order to produce a pH of 2.50?
THANKS!
-J
Let's call propionic acid HP, then it will ionize as
HP ==> H^+ + P^-
First convert pH = 2.50 to (H^+) by
pH = -log(H^+).
Then set up the expression for Ka.
Ka = (H^+)(P^-)/(HP).
You know (H^+) now from the pH calculation. You know (P^-) because it equals the same as (H^+). (HP) = y-(H^+).
Solve for y which = (HP).
Then M x molar mass propanoic acid = grams propanoic acid.
Post your work if you get stuck.
When you do the -log of 2.50 you get a negative number, so I am not really sure what you do with that, do you not use the pKa for anything?
To calculate the mass of propanoic acid needed to produce a pH of 2.50 in a 2.00L volume, we can use the Henderson-Hasselbalch equation.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log ([A-]/[HA])
Where:
- pH is the desired pH (2.50 in this case)
- pKa is the acid dissociation constant of the acid (4.87 for propanoic acid)
- [A-] is the concentration of the conjugate base (propanoate ion)
- [HA] is the concentration of the acid (propanoic acid)
At pH 2.50, we can assume that the concentration of [A-] is negligible compared to [HA]. Thus, the equation simplifies to:
pH ≈ pKa + log ([A-]/[HA])
Rearranging the equation to solve for [A-]/[HA], we get:
[A-]/[HA] ≈ antilog(pH - pKa)
Substituting the values, we have:
[A-]/[HA] ≈ antilog(2.50 - 4.87)
[A-]/[HA] ≈ antilog(-2.37)
[A-]/[HA] ≈ 0.0065
This means that the concentration of the conjugate base is approximately 0.0065 times the concentration of the acid.
Since the volumes are the same (2.00L), we can equate the moles of propanoic acid (HA) to the moles of propanoate ion (A-) using the balanced equation:
HA ⇌ H+ + A-
Assuming complete dissociation, the initial moles of propanoic acid (HA) will be equal to the final moles of propanoate ion (A-).
Let's assume the initial concentration of propanoic acid is C (in moles/L).
Then, the concentration of propanoate ion will be 0.0065C (in moles/L).
The moles of propanoic acid (HA) in 2.0L of solution will be:
Moles of HA = Concentration of HA × Volume of Solution
= C × 2.0
The moles of propanoate ion (A-) in 2.0L of solution will be:
Moles of A- = Concentration of A- × Volume of Solution
= 0.0065C × 2.0
Since the moles of HA is equal to the moles of A-, we can set up an equation:
Moles of HA = Moles of A-
C × 2.0 = 0.0065C × 2.0
Simplifying the equation, we get:
C = 0.0065C
Dividing both sides by 0.0065, we find:
C / 0.0065 = C
This means that the concentration of the propanoic acid should be equal to its original concentration.
To calculate the mass of propanoic acid needed, we can now use the equation:
Mass of propanoic acid = Concentration of propanoic acid × Molar mass × Volume of Solution
The molar mass of propanoic acid (CH3CH2COOH) is 74.10 g/mol.
Substituting the values, we have:
Mass of propanoic acid = C × 74.10 g/mol × 2.0 L
Since we found that C is the concentration of propanoic acid, the mass of propanoic acid needed to produce a pH of 2.50 in a 2.00L volume is given by:
Mass of propanoic acid = (concentration of propanoic acid) × 74.10 g/mol × 2.0 L
Note: Without knowing the initial concentration of propanoic acid, it is not possible to calculate the exact mass needed.
To determine the mass of propanoic acid required to achieve a pH of 2.5, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates pH to the pKa of an acid and the ratio of the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
- pH is the desired pH (2.5 in this case)
- pKa is the negative logarithm of the acid dissociation constant (pKa = 4.87 for propanoic acid)
- [A-] is the concentration of the conjugate base (propanoate ion)
- [HA] is the concentration of the acid (propanoic acid)
In this case, we are looking for the mass of propanoic acid, so we need to know the concentration of propanoic acid. To find this, we need to convert the given volume into moles of propanoic acid.
First, let's convert the given volume of 2.00 liters into grams. We assume that the density of the solution is 1.00 g/mL.
Mass of water = Volume × Density = 2.00 L × 1000 g/L = 2000 g
Now, let's use the molar mass of propanoic acid to convert the mass of water to moles of propanoic acid.
Mass of propanoic acid (in moles) = Mass of water (in grams) / molar mass of propanoic acid
Next, we need to calculate the concentration of propanoic acid (in moles per liter) using the given volume and the moles of propanoic acid calculated above.
Concentration of propanoic acid (in moles/L) = Moles of propanoic acid / Volume of water (in liters)
Now that we have the concentration of propanoic acid, we can substitute the values into the Henderson-Hasselbalch equation and solve for [A-]/[HA].
2.5 = 4.87 + log([A-]/[HA])
Solving for [A-]/[HA] will give us the ratio of the concentrations of propanoate ion to propanoic acid. Let's represent this ratio as "x".
x = [A-]/[HA]
Simplifying the equation:
10^(2.5 - 4.87) = x
x = 10^(-2.37)
Now that we have the ratio of [A-]/[HA], we can calculate the concentration of propanoic acid ([HA]) and the concentration of propanoate ion ([A-]).
Concentration of propanoic acid ([HA]) = Concentration of propanoic acid × (1 / (1 + x))
Concentration of propanoate ion ([A-]) = Concentration of propanoic acid × (x / (1 + x))
Finally, we multiply the concentration of propanoic acid by the volume to get the mass of propanoic acid required.
Mass of propanoic acid = Concentration of propanoic acid × Volume
After substituting the known values and solving for the mass of propanoic acid, you will get the answer.
Note: Please double-check the calculations for accuracy.