A 0.035 M solution of a weak acid (HA) has a pH of 4.88. What is the Ka of the acid?
HA ==>H^+ + A^-
Ka = (H^+)(A^-)/(HA)
pH = -log(H^+)
You know pH. Convert that to (H^+).
(H^+)=(A^-) so plug those into the expression for Ka.
(HA) = 0.035 M - (H^+). Plug that in.
Calculate Ka.
Post your work if you get stuck.
THANK YOU SOO MUCH!!! I don't know why I cant get the simplest problems!
Well, here's a simple solution for you! Let's break it down step by step:
Step 1: Calculate the concentration of H+ ions.
Since the pH is 4.88, the concentration of H+ ions can be calculated using the formula: [H+] = 10^(-pH). Plugging in the value of pH = 4.88, we get [H+] = 10^(-4.88).
Step 2: Calculate the concentration of HA.
Since the concentration of HA is given as 0.035 M and (HA) = 0.035 M - [H+], we can substitute the value of [H+] we calculated in step 1 to find the concentration of HA.
Step 3: Calculate Ka.
Using the formula for Ka = (H+)(A-)/(HA), we can substitute the values of [H+] and (HA) we calculated in steps 1 and 2 to find the Ka value.
Now, give it a try and see if you can find the answer! I'll be here to assist if you get stuck.
No worries! I can help you with the problem step-by-step. Let's start with the information given:
- The pH of the solution is 4.88.
- The concentration of the weak acid (HA) is 0.035 M.
Step 1: Convert the pH to the concentration of H^+ ions.
The pH is given as 4.88. To convert this to the concentration of H^+ ions, we need to use the formula:
pH = -log[H+]
Let's plug in the given pH value and solve for [H+]:
4.88 = -log[H+]
Taking the antilog of both sides, we get:
[H+] = 10^(-pH)
[H+] = 10^(-4.88)
[H+] ≈ 1.27 × 10^(-5) M
So, the concentration of H^+ ions is approximately 1.27 × 10^(-5) M.
Step 2: Calculate the concentration of A^- ions.
In the dissociation of the weak acid HA, the concentration of A^- ions is equal to the concentration of H^+ ions because the acid is monoprotic.
Therefore, the concentration of A^- ions is also approximately 1.27 × 10^(-5) M.
Step 3: Calculate the concentration of the weak acid HA.
The concentration of HA can be calculated by subtracting the concentration of H^+ ions from the initial concentration of the weak acid, which is given as 0.035 M:
(HA) = 0.035 M - (1.27 × 10^(-5) M)
(HA) ≈ 0.0349873 M
Step 4: Calculate the Ka of the acid.
Now that we have all the required concentrations, we can use the Ka expression:
Ka = (H+)(A^-)/(HA)
Substituting the values:
Ka = (1.27 × 10^(-5) M)(1.27 × 10^(-5) M)/(0.0349873 M)
Calculating the value, we get:
Ka ≈ 4.6 × 10^(-10)
Therefore, the Ka of the weak acid is approximately 4.6 × 10^(-10).
To solve this problem, you can follow these steps:
Step 1: Convert pH to (H+)
Use the equation pH = -log(H+), where pH = 4.88.
Take the antilog of both sides to get rid of the negative sign:
H+ = 10^(-pH)
Step 2: Calculate (HA) concentration
Given that (HA) is a 0.035 M solution, subtract the (H+) concentration from the total concentration to get (HA):
(HA) = 0.035 M - (H+)
Step 3: Calculate Ka
Use the equation Ka = (H+)(A-)/(HA).
Since (H+) equals (A-), substitute them into the equation:
Ka = [(H+)(H+)]/(HA)
Ka = (H+)^2/(HA)
Step 4: Substitute values and calculate Ka
Plug in the values you obtained in steps 1 and 2 into the equation for Ka:
Ka = [(10^(-pH))(10^(-pH))]/(0.035 M - (10^(-pH)))
Ka = (10^(-2pH))/(0.035 M - (10^(-pH)))
Using a calculator, substitute the value of pH = 4.88 into these equations to calculate (H+), (HA), and eventually Ka.
If you encounter any issues or need further assistance, feel free to ask!