How many grams of benzoic acid, a monoprotic acid with a Ka=6.4x10 to the negative 5, must be dissolved in 500.0 mL of water to produce a solution with a pH=2.50?
0.156 is what I have also. No, don't divide by 0.5.
M = mols/L. If you want 0.156 molar in 0.5 L, then you want 1/2 that amount in 500 mL or 0.5 x 0.156 = 0.078 mols in 500 mL.
mols = g/molar mass
0.078 mols = grams/molar mass.
You know molar mass, calculate grams.
HBz ==> H^+ + Bz^-
Ka = [(H^+)(Bz^-)]/(HBz)
Use pH = 2.50 = - log(H^+) to determine (H^+). Set that in the Ka expression. (Bz^-) is the same as (H^+).
(HBz) = unknown. Solve for that to determine molarity. Change to mols/500 mL, then to grams. Post your work if you get stuck.
Ok, so I'm at 1.56x10^-1 M, would I divide that by .5 L right away?
To determine the number of grams of benzoic acid needed, we can follow these steps:
Step 1: Convert the given pH to the concentration of H₃O⁺ ions.
- The pH is a measure of the concentration of H₃O⁺ ions in a solution.
- The concentration of H₃O⁺ ions can be determined using the formula:
[H₃O⁺] = 10^(-pH)
- In this case, [H₃O⁺] = 10^(-2.50) = 0.00316 M
Step 2: Write the balanced equation for the dissociation of benzoic acid.
- Benzoic acid (C₆H₅COOH) dissociates in water to produce H₃O⁺ and the conjugate base, C₆H₅COO⁻.
- The balanced equation is: C₆H₅COOH ⇌ H₃O⁺ + C₆H₅COO⁻
Step 3: Write the expression for the acid dissociation constant (Ka).
- The Ka expression for benzoic acid is: Ka = [H₃O⁺][C₆H₅COO⁻] / [C₆H₅COOH]
Step 4: Express the concentrations in terms of x (the change in concentration during dissociation).
- [H₃O⁺] = x
- [C₆H₅COO⁻] = x
- [C₆H₅COOH] = initial concentration - x
Step 5: Plug the expressions into the Ka expression and solve for x.
- Ka = [H₃O⁺][C₆H₅COO⁻] / [C₆H₅COOH]
- 6.4x10^(-5) = x * x / (initial concentration - x)
Step 6: Solve for x using the quadratic equation and approximation.
- Rearrange the equation to: x^2 - 1.6x10^(-5)x - 6.4x10^(-5) = 0
- Solve the quadratic equation or use approximation methods to find x.
Step 7: Calculate the moles of benzoic acid using the concentration.
- moles = concentration * volume
- The volume is given as 500.0 mL, which is equivalent to 0.5000 L.
- moles of benzoic acid = concentration * volume = x * 0.5000
Step 8: Convert moles of benzoic acid to grams.
- Convert moles to grams using the molar mass of benzoic acid (C₆H₅COOH).
- The molar mass of C₆H₅COOH = (6 x Atomic mass of C) + (5 x Atomic mass of H) + Atomic mass of O
- Calculate the molar mass and then multiply it by the number of moles of benzoic acid obtained in Step 7.
By following these steps, you should be able to calculate the number of grams of benzoic acid needed to produce the desired solution.