How many grams of NAC2H3O2 must be dissolve in enough water to prepare 250.0 mL of a solution with a pH= 9.200? Molar mass of NaC2H3O2 ia 82.03 g/mol. Ka for C2H3O2H is 1.8x10^-5
If we call acetic acid HAc, then the pH will be determined by the hydrolysis of the salt. Convert pH 9.200 to OH^-. I get about 1.6E-5M
..............Ac^- + HOH -->HAc + OH^-
initial......x..............0.....0
equil......x-1.6E-5.......1.5E-5..1.6E-5
Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
Substitute and solve for Ac^-,(answer will be in moles/L) and multiply that by 0.250 to find moles in 250 mL, then
that x molar mass NaAc give grams NaAc in 250 mL.
omg, thank u! I ended up getting something 1.70! THANKS!
Where I wrote 1.6E-5 for concn OH^-, I should have typed 1.6E-5 ALSO for (HAc) but I made a typo with 1.5E-5. By the way,
I think the answer is closer to 9 g. Post your work if you wish and I will look for the error.
I actually just redid it and found something close to 9.21?
Right. My answer was 9.22 g.
To solve this problem, we need to use the Henderson-Hasselbalch equation, which relates the pH of a solution to the acid dissociation constant (Ka) and the concentrations of the acid and its conjugate base.
The Henderson-Hasselbalch equation for this problem is:
pH = pKa + log ([A-]/[HA])
Where:
- pH is the given pH value, which is 9.200 in this case.
- pKa is the negative logarithm of the acid dissociation constant (Ka), which is given as 1.8x10^-5.
- [A-] is the concentration of the conjugate base, which is the acetate ion (C2H3O2-).
- [HA] is the concentration of the acid, which is acetic acid (C2H3O2H) in this case.
First, let's calculate the concentration of acetic acid ([HA]). We'll assume that the sodium acetate (NaC2H3O2) fully dissociates in water, so the concentration of C2H3O2- will be equal to [NaC2H3O2].
Given that the molar mass of NaC2H3O2 is 82.03 g/mol, we need to convert the volume of the solution to liters:
250.0 mL = 250.0 mL × (1 L / 1000 mL) = 0.250 L
Next, we need to calculate the moles of NaC2H3O2 using its molar mass:
moles of NaC2H3O2 = mass of NaC2H3O2 / molar mass of NaC2H3O2
moles of NaC2H3O2 = (mass of NaC2H3O2) / 82.03 g/mol
To find the mass of NaC2H3O2, we need to rearrange the Henderson-Hasselbalch equation to solve for [A-]:
[A-] = 10^(pH - pKa) × [HA]
Substituting the given values, we have:
[A-] = 10^(9.200 - (-5)) × [HA]
Now, we substitute [A-] and [HA] with their respective concentrations and solve for the mass of NaC2H3O2:
(moles of NaC2H3O2) / 0.250 L = 10^(9.200 - (-5))
mass of NaC2H3O2 = (moles of NaC2H3O2) × molar mass of NaC2H3O2
Finally, we substitute the values and calculate the mass of NaC2H3O2:
mass of NaC2H3O2 = (moles of NaC2H3O2) × molar mass of NaC2H3O2
So, to find the mass of NaC2H3O2, you will need to:
1. Convert the volume of the solution to liters.
2. Calculate the moles of NaC2H3O2.
3. Use the Henderson-Hasselbalch equation to find the concentration of [A-].
4. Substitute the concentrations of [A-] and [HA] into the equation to find the mass of NaC2H3O2.