When 0.285 g of a solid weak acid HA (molar mass = 184 g mol−1) is dissolved in water to a total volume of 25.0 mL, the pH of the solution is 3.59.
What is the acid ionization constant (Ka) of the acid?
will it be Ka=HA/gxvol?
I don't think it's quite that simple.
(HA) = mols/L = g/molar mass/L = 0.285/184/0.025 = about 0.06.
................HA ==> H^+ + A^-
I...............0.06.......0..........0
C...............-x..........x...........x
E............0.06-x......x............x
The problem tells you pH = 3.59 so 3.59 = -log(H^+) and (H^+) = 2.57E-4 M.
Plug the E line into the Ka expression and you get (x)(x)/(0.06-x) = Ka.
You know what x is; i.e., 2.57E-4 so plug that in and solve for Ka.
To find the acid ionization constant (Ka) of the acid, we can use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a solution to the pKa (the negative logarithm of Ka) and the ratio of the concentration of the conjugate base to the concentration of the acid.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
In this case, we are given the pH (3.59) and the concentration ([HA]) of the acid.
To find [A-], we can consider that the total amount of acid, HA, is equal to the sum of the acid that is ionized, [A-], and the acid that is not ionized, [HA].
[HA] + [A-] = 0.285 g
To convert the mass of the acid into moles, we can use the molar mass of the acid:
moles of acid (HA) = mass (g) / molar mass (g/mol)
moles of acid (HA) = 0.285 g / 184 g/mol
Now, we can find the concentration of the acid ([HA]) by dividing the moles of acid by the volume of the solution:
[HA] = moles of acid (HA) / volume of solution (L)
[HA] = (0.285 g / 184 g/mol) / 0.025 L
Now that we have the concentration of the acid ([HA]) and the pH, we can rearrange the Henderson-Hasselbalch equation to solve for pKa:
pKa = pH - log([A-]/[HA])
pKa = 3.59 - log([A-] / 0.285 g / 184 g/mol / 0.025 L)
Simplifying,
pKa = 3.59 - log([A-] / 1.37 mol/L)
To find Ka, we can take the antilog of -pKa:
Ka = 10^(-pKa)
Therefore, to find the acid ionization constant (Ka) of the acid, we need to calculate [A-] using the molecular weight (molar mass) of the acid, and then substitute the values into the equation above.
To find the acid ionization constant (Ka) of the acid, we first need to determine the concentration of the weak acid HA in the solution. We can do this by using the given mass (0.285 g) and molar mass (184 g mol−1).
First, calculate the moles of HA:
moles of HA = mass of HA / molar mass of HA
moles of HA = 0.285 g / 184 g mol−1
Next, we need to find the concentration of HA in mol L−1. We can do this by dividing the moles of HA by the volume of the solution in liters.
Volume of solution = 25.0 mL = 25.0 mL / 1000 mL L−1 = 0.025 L
Concentration of HA = moles of HA / volume of solution
Concentration of HA = (0.285 g / 184 g mol−1) / 0.025 L
Now that we have the concentration of HA, we can find the concentration of H+ ions in the solution. Since the pH of the solution is 3.59, we know that the concentration of H+ ions is 10^(-pH).
Concentration of H+ = 10^(-pH)
Concentration of H+ = 10^(-3.59)
The acid dissociation reaction can be written as:
HA + H2O ⇌ H3O+ + A-
The initial concentration of HA is the same as the concentration of HA we calculated earlier. Let's assume the concentration of H3O+ and A- as x (since they have the same molar concentration due to 1:1 stoichiometry).
Using the given equation for Ka:
Ka = (H3O+ * A-) / HA
Substituting the values we have obtained:
Ka = (x * x) / (concentration of HA)
Now, we can put the given pH value into the concentration of H+ equation to solve for x:
10^(-3.59) = x / (concentration of HA)
By solving this equation, we can find the value of x.
Finally, substitute the value of x into the equation for Ka to obtain the acid ionization constant of the acid (HA).