A buffer solution is prepared by adding 14.37g of NaC2H3O2 and 16.00g of acetic acid to enough water to make 500mL of solution.
What is the initial concentration of C2H3O2 in the solution?
Ka(HC2H3O2)=1.8e-5
HC2H3O2 = H3O+ + c2h3O2
the molar mass of NaC2H3O3 = 82.04
I know that m=mol NaC2H3O2 / L
so would I do 82.04 / .05?
You don't know that m = mol NaC2H3O2/L; because m stands for molality. You may know that M = mol NaC2H3O2/L since that is molarity.
mols NaC2H3O2 = grams/molar mass
mols HC2H3O2 = grams/molar mass
M of each = mols/0.500 L
Then pH = pKa + log(base)/(acid).
To find the initial concentration of C2H3O2 in the solution, you need to calculate the number of moles of NaC2H3O2 in the solution. Here's how you can do it step by step:
1. Calculate the moles of NaC2H3O2:
- Given mass of NaC2H3O2 = 14.37g
- Molar mass of NaC2H3O2 = 82.04 g/mol
- Moles of NaC2H3O2 = mass / molar mass
= 14.37g / 82.04 g/mol
2. Calculate the moles of acetic acid (HC2H3O2):
- Given mass of acetic acid = 16.00g
- Molar mass of HC2H3O2 = 60.05 g/mol
- Moles of HC2H3O2 = mass / molar mass
= 16.00g / 60.05 g/mol
3. Add the moles of NaC2H3O2 and HC2H3O2 to get the total moles of C2H3O2 in the solution.
4. Convert the moles of C2H3O2 to the initial concentration:
- Given volume of solution = 500mL = 0.5L
- Initial concentration (molarity) of C2H3O2 = moles of C2H3O2 / volume of solution
So, to answer your question, you would calculate the moles of NaC2H3O2 and HC2H3O2 separately, add them together, and then divide by the volume of the solution.