What mass of benzoic acid CH6H5COOH, would you dissolve in 400.0 mL of water to produce a solution with a pH= 2.90? Ka= 6.3 x 10^-5
mass in grams is needed.
Let HB = benzoic acid.
HB ==> H^+ + B^-
Ka = (H^+)(B^-)/(HB)
Set up an ICE chart.
Convert pH=2.90 to (H^+). Substitute into Ka expression for (H^+) and for (B^-) (they are equal). Solve for (HB) which will be in units of M = moles/L. You know M, you know how many L you want (0.4000), solve for moles. Then moles = grams/molar mass and solve for grams.
To find the mass of benzoic acid needed to produce a solution with a pH of 2.90, we need to use the equation for the dissociation of benzoic acid, as well as the ionization constant (Ka) of benzoic acid.
The dissociation equation for benzoic acid in water is:
CH6H5COOH ⇌ CH6H5COO- + H+
The Ka expression for this reaction is:
Ka = [CH6H5COO-][H+] / [CH6H5COOH]
Given that the Ka value is 6.3 x 10^-5, we can set up the equation for the dissociation of benzoic acid as follows:
6.3 x 10^-5 = [CH6H5COO-][H+] / [CH6H5COOH]
Since we want to know the mass of benzoic acid needed to achieve a pH of 2.90, we can assume that the initial concentration of benzoic acid is equal to the final concentration of benzoic acid.
Let's assume the concentration of benzoic acid in the solution is "x" mol/L. Since the volume of the solution is given as 400.0 mL, we can convert it to liters by dividing by 1000:
Volume (L) = 400.0 mL / 1000 = 0.400 L
Now, we can substitute the known values into the Ka expression:
6.3 x 10^-5 = (x)(x) / (0.400 - x)
We can neglect the value of "x" in the denominator because it will be very small compared to 0.400. Therefore, our equation becomes:
6.3 x 10^-5 = x^2 / 0.400
Rearranging the equation, we get:
x^2 = 6.3 x 10^-5 * 0.400
x^2 = 2.52 x 10^-5
Taking the square root of both sides, we find:
x = √(2.52 x 10^-5) = 0.00502 mol/L
Now, we can calculate the molar mass of benzoic acid (C7H6O2):
Molar mass of benzoic acid = (7 x atomic mass of carbon) + (6 x atomic mass of hydrogen) + (2 x atomic mass of oxygen)
Using the atomic masses from the periodic table, we find:
Molar mass of benzoic acid ≈ (7 x 12.01 g/mol) + (6 x 1.01 g/mol) + (2 x 16.00 g/mol)
Molar mass of benzoic acid ≈ 84.11 g/mol
Finally, we can calculate the mass of benzoic acid needed by multiplying the concentration (in mol/L) by the volume (in L) and molar mass (in g/mol):
mass of benzoic acid = 0.00502 mol/L * 0.400 L * 84.11 g/mol
mass of benzoic acid ≈ 0.168 g
Therefore, approximately 0.168 grams of benzoic acid would need to be dissolved in 400.0 mL of water to produce a solution with a pH of 2.90.