Find the pH of a Sol that is 0.15 M HCOOH (Ka = 1.8e-4) and 0.45 M in HCN (KA = 4.9e-10).
Options are 3.87, 2.28, 4.38, 3.05, 4.13
I only that we use handerson eq but there’s two Ka given.
Answers for AP Chemistry Partner Quiz (I have a PDF of the answer key if you don't fully believe me)
1. E
2. C
3. D
4. C
5. B
6. A
7. C
8. D
9. A
10. A
11. D
12. B
13. E
14. C
15. B
16. D
17. C
18. E
19. D
20. D
21. A
22. A
23. B
24. E
25. B
Which question is that for?
Who suggested that you should use the Henderson-Hasselbalch equation? I'm not sure that is the proper approach. If you still need help I can show you how to do it. By the way, since you have received some unasked for answers from another, please go to a post just above yours from JOHN and read a comment I made to that. You want to learn how to use spurious answers properly.
To find the pH of the solution, we can use the Henderson-Hasselbalch equation. However, since there are two different acids (HCOOH and HCN) present in the solution, we need to consider the equilibrium reactions for both acids separately.
1. For HCOOH:
HCOOH ⇌ H⁺ + COO⁻
Let x be the concentration of H⁺ ions produced from HCOOH. Since the initial concentration of HCOOH is 0.15 M, the concentration of HCOOH remaining will be (0.15 - x) M, and the concentration of COO⁻ will also be x M.
Using the given Ka value of HCOOH, we can set up the following equilibrium expression:
Ka = [H⁺][COO⁻] / [HCOOH]
1.8e-4 = x * x / (0.15 - x)
Assuming that x is much smaller than 0.15, we can approximately say (0.15 - x) ≈ 0.15. Therefore, the equation can be simplified to:
1.8e-4 = x^2 / 0.15
Rearranging the equation gives:
x^2 = 1.8e-4 * 0.15
x = sqrt(1.8e-4 * 0.15)
x ≈ 0.0195 M
The concentration of H⁺ ions from HCOOH is approximately 0.0195 M.
2. For HCN:
HCN ⇌ H⁺ + CN⁻
Let y be the concentration of H⁺ ions produced from HCN. Since the initial concentration of HCN is 0.45 M, the concentration of HCN remaining will be (0.45 - y) M, and the concentration of CN⁻ will also be y M.
Using the given Ka value of HCN, we can set up the following equilibrium expression:
Ka = [H⁺][CN⁻] / [HCN]
4.9e-10 = y * y / (0.45 - y)
Assuming that y is much smaller than 0.45, we can approximately say (0.45 - y) ≈ 0.45. Therefore, the equation can be simplified to:
4.9e-10 = y^2 / 0.45
Rearranging the equation gives:
y^2 = 4.9e-10 * 0.45
y = sqrt(4.9e-10 * 0.45)
y ≈ 3.83e-6 M
The concentration of H⁺ ions from HCN is approximately 3.83e-6 M.
Now that we have the concentrations of H⁺ ions from both acids, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A⁻]/[HA])
For HCOOH:
pKa = -log(Ka) = -log(1.8e-4) ≈ 3.74
[HA] = 0.15 - x ≈ 0.15 - 0.0195 ≈ 0.1305 M
[A⁻] = x ≈ 0.0195 M
For HCN:
pKa = -log(Ka) = -log(4.9e-10) ≈ 9.31
[HA] = 0.45 - y ≈ 0.45 - 3.83e-6 ≈ 0.449996 M
[A⁻] = y ≈ 3.83e-6 M
Substituting the values into the equation:
pH = 3.74 + log (0.0195/0.1305) + 9.31 + log (3.83e-6/0.449996)
pH = 3.74 + (-1.16) + 9.31 + (-5.82)
Calculating the sum:
pH ≈ 2.07
Therefore, the pH of the solution is approximately 2.07.
None of the provided answer choices match the calculated pH. Please double-check the options or input to ensure accuracy.