Question

Calculate the pH that results when the following solutions are mixed

1. 35 mL of 0.20 M formic acid 𝐻𝐶𝑂𝑂𝐻, 𝐾𝑎 = 1.77 × 10−4
2. 55 mL of 0.10 M sodium formate 𝐻𝐶𝑂𝑂𝑁𝑎
3. 110 mL of water

Answer 1

Use the HH equation.

1. 35 mL of 0.20 M formic acid 𝐻𝐶𝑂𝑂𝐻, 𝐾𝑎 = 1.77 × 10−4
2. 55 mL of 0.10 M sodium formate 𝐻𝐶𝑂𝑂𝑁𝑎
3. 110 mL of water
Total volume is 200 mL.
(HCOOH) = 0.20 M x (35/200) = ?
(HCOONa) = 0.10 M x (55/200).
Convert Ka to pKa.Substitute into the HH equation. Solve for pH.
Note that HCOOH is the acid and HCOONa is the base.
Post your work if you get stuck.

Answer 2

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Answer 3

Calculate the pH that results when the following solutions are mixed

1. 35 mL of 0.20 M formic acid 𝐻𝐶𝑂𝑂𝐻, 𝐾𝑎 = 1.77 × 10−4
2. 55 mL of 0.10 M sodium formate 𝐻𝐶𝑂𝑂𝑁𝑎
3. 110 mL of water

Answer 4

[AI-generated answer — Clown Bot]

Well, when it comes to mixing solutions, it can get a little tricky. But fear not, I'm here to clown around with your question!

To determine the pH of this mixture, we'll have to look at the relationship between formic acid and its conjugate base, sodium formate. You see, formic acid (HCOOH) can donate a proton (H+) while sodium formate (HCOONa) can accept a proton (H+). This means that when formic acid reacts with sodium formate, some of the formic acid will lose its protons to the sodium formate.

So, in this case, we'll have a mixture of formic acid and its conjugate base. And since formic acid is a weak acid, it will partially dissociate in water.

But calculating the pH can be as fun as juggling rubber chickens! To determine the resulting pH, we need to calculate the concentration of the formic acid and sodium formate in the final solution.

First, let's calculate the mols of formic acid and sodium formate in their respective solutions:
Mols of formic acid = (35 mL * 0.20 M) = 7 mmol
Mols of sodium formate = (55 mL * 0.10 M) = 5.5 mmol

Now, let's calculate the resulting concentrations of formic acid and sodium formate in the mixture. Since the total volume of the mixture is 35 mL + 55 mL + 110 mL = 200 mL:

Concentration of formic acid in the mixture = (7 mmol / 200 mL) = 0.035 M
Concentration of sodium formate in the mixture = (5.5 mmol / 200 mL) = 0.0275 M

Now, we need to determine the ratio of the concentrations of the formate ion (HCOO-) to formic acid (HCOOH). In this case, it will be 0.0275 M / 0.035 M = 0.7857.

To determine the pH, we can use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-] / [HA]),

where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

For formic acid, pKa = -log(1.77 × 10^-4) = 3.75.

Plugging in the values, we get:

pH = 3.75 + log (0.0275 M / 0.035 M)
pH = 3.75 + log (0.7857)

And if we plug that into our imaginary clown calculator, we find that the resulting pH is approximately 3.20.

So, the pH of this mixture is approximately 3.20. Tada! I hope I brought a smile to your face with my clownish calculations!

Answer 5

[AI-generated answer — Explain Bot]

To calculate the pH of the resulting mixture, we need to consider the acid-base reaction that occurs when formic acid (HCOOH) and sodium formate (HCOONa) are mixed.

The chemical equation for the reaction between formic acid and sodium formate can be written as:

HC6H7O2 + NaOH -> H2O + NaC6H7O2

In this reaction, formic acid (HC6H7O2) acts as an acid, and sodium formate (NaC6H7O2) acts as a base. The reaction produces water (H2O) and sodium formate (NaC6H7O2).

To calculate the pH, we need to determine the concentrations of the resulting solution's components.

Step 1: Calculate the moles of formic acid and sodium formate.
- Moles of formic acid (HC6H7O2) = volume (L) x concentration (mol/L) = 0.035 L x 0.20 mol/L = 0.007 mol
- Moles of sodium formate (NaC6H7O2) = volume (L) x concentration (mol/L) = 0.055 L x 0.10 mol/L = 0.0055 mol

Step 2: Determine the limiting reactant.
To identify the limiting reactant, we need to compare the number of moles of the reactants. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed. In this case, we can see that formic acid (HC6H7O2) is present in fewer moles than sodium formate (NaC6H7O2). Therefore, formic acid is the limiting reactant.

Step 3: Calculate the concentration of the excess reagent (sodium formate).
Since formic acid is the limiting reactant, there will be some excess sodium formate left after the reaction. To calculate the concentration of the excess reagent, we need to subtract the moles of formic acid that reacted from the total moles of sodium formate.
- Moles of sodium formate remaining = moles of sodium formate - moles of formic acid reacted
= 0.0055 mol - 0.007 mol = -0.0015 mol (negative value indicates excess)

Step 4: Calculate the concentration of the resulting solution.
The resulting solution is a mixture of the products of the reaction: water (H2O) and sodium formate (NaC6H7O2). Since water is a neutral molecule, it does not affect the pH. Therefore, we need to calculate the concentration of sodium formate.

- Volume of the resulting solution = total volume of formic acid + total volume of sodium formate + total volume of water
= 0.035 L + 0.055 L + 0.11 L = 0.20 L

- Concentration of sodium formate in the resulting solution = moles of sodium formate remaining / volume of the resulting solution
= (-0.0015 mol) / (0.20 L) = -0.0075 M (negative value indicates excess)

Step 5: Calculate the pOH of the resulting solution.
The pOH is calculated by taking the negative logarithm (base 10) of the hydroxide ion concentration. In this case, since we are dealing with the base sodium formate, we can calculate the pOH from the concentration of sodium formate.

pOH = -log10([OH-]) = -log10(0.0075 M) ≈ 2.12

Step 6: Calculate the pH of the resulting solution.
The pH of the resulting solution can be calculated using the equation:

pH + pOH = 14

pH = 14 - pOH = 14 - 2.12 ≈ 11.88

Therefore, the pH of the resulting solution when the three solutions are mixed is approximately 11.88.