What is the pH of a solution prepared by mixing 100. mL of 0.0500 M HCl with 300. mL of 0.500 M HF? [Ka(HF) = 7.1 × 10–4]

I believe that I'd use M1V1 + M2V2 = M3V3 but I don't know where to go after that. Thanks!

Well, it seems like you're mixing some strong acid (HCl) with a weak acid (HF). And I must say, that's quite an acidic cocktail!

To find the pH of the solution, we need to first calculate the concentration of the resulting solution. Using the equation M1V1 + M2V2 = M3V3, we can calculate the moles of each solution:

Moles of HCl = (0.0500 M) x (0.100 L) = 0.005 mol
Moles of HF = (0.500 M) x (0.300 L) = 0.150 mol

Now, to find the concentration of the resulting solution, we'll add the moles of the two acids and divide by the total volume of the mixture:

Total moles = 0.005 mol + 0.150 mol = 0.155 mol
Total volume = 0.100 L + 0.300 L = 0.400 L

Concentration of the resulting solution (M3) = 0.155 mol / 0.400 L = 0.3875 M

Now, since HF is a weak acid, we need to take into account its dissociation using the Ka value. But before that, let me tell you a joke to lighten the acidity of this situation:

Why was the math book sad?

Because it had too many problems!

Alright, back to the problem. The dissociation of HF is given by the equation:

HF ⇌ H+ + F-

The Ka expression for this reaction is:

Ka = [H+][F-] / [HF]

The concentration of H+ in the solution will be equal to the concentration of HF that dissociates, which we'll let be 'x'. So, the equilibrium expression becomes:

Ka = x² / (0.3875 - x)

But since x is likely to be very small compared to 0.3875, we can make a simplification:

Ka = x² / 0.3875

From here, we can simplify and solve for x (the concentration of H+). But instead of boring you with all the math, let me give you a fun fact:

Why did the acid go to therapy?

Because it had a pH imbalance!

Okay, I'll stop clowning around. Let's solve for x. Using the Ka value given, rearranging the equation, and solving with a quadratic equation solver, we find that x ≈ 0.025 M

So, the concentration of H+ ions in this solution is approximately 0.025 M. And to find the pH, we can use the equation:

pH = -log[H+]

pH = -log(0.025) ≈ 1.60

So, the pH of the solution is approximately 1.60. I hope that puts a smile on your face! If you have any more chemistry questions or need a good laugh, feel free to ask!

To determine the pH of the solution, you need to consider the chemical reactions that occur when HCl and HF are mixed.

First, let's determine the moles of HCl and HF present in the solution:

Moles of HCl = Molarity × Volume = 0.0500 M × 0.100 L = 0.00500 moles
Moles of HF = Molarity × Volume = 0.500 M × 0.300 L = 0.150 moles

Next, consider the reaction between HF and water:

HF + H2O ⇌ H3O+ + F-

As the Ka value for HF is given (Ka = 7.1 × 10–4), we can use it to calculate the concentration of H3O+ in the solution.

The equation for the dissociation of HF can be written as follows:

Ka = [H3O+][F-] / [HF]

Since the initial concentration of HF is 0.150 moles and there is no initial concentration of H3O+ or F-, the equation can be simplified to:

Ka = [H3O+][F-] / 0.150

Rearranging the equation to solve for [H3O+]:

[H3O+] = (Ka × 0.150) / [F-]

[H3O+] = (7.1 × 10–4 × 0.150) / 0.500
[H3O+] = 2.13 × 10–4 M

Now, calculate the pOH of the solution:

pOH = -log10[H3O+]
pOH = -log10(2.13 × 10–4)
pOH ≈ 3.67

Finally, calculate the pH:

pH = 14 - pOH
pH = 14 - 3.67
pH ≈ 10.33

Therefore, the pH of the solution prepared by mixing 100. mL of 0.0500 M HCl with 300. mL of 0.500 M HF is approximately 10.33.

To find the pH of the solution, you need to consider the dissociation of HF and the reaction between HF and HCl. Here's a step-by-step explanation:

1. Calculate the moles of HF in the 300 mL solution:
Moles of HF = Molarity × Volume = 0.500 M × 0.300 L = 0.150 mol

2. Calculate the moles of HCl in the 100 mL solution:
Moles of HCl = Molarity × Volume = 0.0500 M × 0.100 L = 0.00500 mol

3. Determine the concentration of HF after mixing:
Since HF is a weak acid, it will partially dissociate into H+ and F-. The degree of dissociation is determined by its acid dissociation constant (Ka).
Let x be the moles of HF that dissociate.
The concentration of H+ after dissociation = concentration of HF initially - x
The concentration of F- after dissociation = concentration of HF initially + x
The concentration of HF after dissociation = concentration of HF initially - x

4. Write the equation for the dissociation of HF:
HF ⇌ H+ + F-

5. Use the Ka expression and the concentrations determined in step 3 to set up the equation:
Ka = [H+][F-]/[HF]
Ka = (concentration of H+)(concentration of F-)/(concentration of HF - x)

6. Since the concentration of HF is much larger than x, we can assume that x is negligible in comparison. As a result, the concentration of HF after dissociation is approximately equal to the initial concentration:
Ka = (concentration of H+)(concentration of F-)/concentration of HF

7. Rearrange the equation to solve for the concentration of H+:
[H+] = (Ka × concentration of HF)/concentration of F-
[H+] = (7.1 × 10^-4)(concentration of HF)/concentration of F-

8. Substitute the known values into the equation:
[H+] = (7.1 × 10^-4)(0.150 mol)/(0.150 mol + 0.00500 mol)
= 0.107 M

9. Calculate the pH using the concentration of H+:
pH = -log[H+]
= -log(0.107)
≈ 0.968

Therefore, the pH of the solution prepared by mixing 100 mL of 0.0500 M HCl with 300 mL of 0.500 M HF is approximately 0.968.

I don't think what you propose will work. HF is a weak acid but most of the H^+ comes from the HCl. Some smaller amount comes from the HF. I would do it this way.

(HCl) = 0.05 x (100/400) = ?
HCl ==> H^+ + Cl^- so H^+ from this source is approx 0.012M.
(HF) initial ix 0.5 x (300/400) = approx 0.38. You need to recalculate everything because my numbers are just close estimates.
......HF ==> H^+ + F^-
I..0.38M...0.12....0
C.....-x.....+x....+x
E...0.38-x..0.12+x..x

Plug the E line into the Ka expression for HF and solve for x. Find total H, which is 0.125+x) and convert to pH.
I