Calculate the pH after 0.10 mol of HCl is added to 1.00 L of the solution that is 0.57 M HF and 1.13 M KF.

Question

Calculate the pH after 0.10 mol of HCl is added to 1.00 L of the solution that is 0.57 M HF and 1.13 M KF.

I got 3.01, but it is wrong. What did I do wrong?

Answer 1

To calculate the pH after adding HCl to a solution of HF and KF, you need to consider the ionization of HF and the reaction between HF and HCl. Let's break down the steps to solve this problem:

Step 1: Calculate the initial concentration of HF (hydrofluoric acid) and KF (potassium fluoride) in moles per liter (M).

Given:
- Initial volume (V): 1.00 L
- Concentration of HF: 0.57 M
- Concentration of KF: 1.13 M

So, moles of HF (nHF) = Concentration of HF (CHF) x Volume (V) = 0.57 M x 1.00 L = 0.57 mol
And moles of KF (nKF) = Concentration of KF (CKF) x Volume (V) = 1.13 M x 1.00 L = 1.13 mol

Step 2: Determine the reaction between HF and HCl.

HF (aq) + H2O (l) ⇌ H3O+ (aq) + F- (aq)

In this reaction, HF acts as a weak acid that partially ionizes in water. So, the concentration of H3O+ and F- will depend on the equilibrium constant (Ka) of this reaction.

Step 3: Calculate the new concentrations of HF, H3O+, and F- after the addition of HCl.

Given:
- Moles of HCl added: 0.10 mol
- Initial concentration of HF: 0.57 M

The reaction between HF and HCl can be summarized as:

HF (aq) + HCl (aq) ⇌ H3O+ (aq) + Cl- (aq)

Since HF is a weak acid and HCl is a strong acid, the reaction will favor the formation of the products. The moles of HF remaining after the reaction will be nHF - nHCl = 0.57 mol - 0.10 mol = 0.47 mol.

The moles of H3O+ formed will be equal to the moles of HCl added, so nH3O+ = 0.10 mol.

Step 4: Determine the new concentrations of HF, H3O+, and F-.

New concentration of HF (CHF_new) = moles of HF remaining / volume = 0.47 mol / 1.00 L = 0.47 M
New concentration of H3O+ (CH3O+_new) = moles of H3O+ formed / volume = 0.10 mol / 1.00 L = 0.10 M
New concentration of F- (CF-_new) = concentration of KF = 1.13 M

Step 5: Calculate the pH of the solution.

To calculate the pH, we need to determine the concentration of H3O+ (CH3O+) in the final solution. Since it is a strong acid (HCl), we can assume that it fully dissociates in water.

Thus, the concentration of H3O+ is equal to the concentration of HCl added:

CH3O+ = 0.10 M

Now, we can calculate the pH using the formula:

pH = -log10(CH3O+)

pH = -log10(0.10) = 1

So, the correct pH after adding 0.10 mol of HCl is 1, not 3.01.

You mistakenly calculated the pH based on the concentration of HCl added instead of the concentration of H3O+ formed. Make sure to use the correct value to calculate the pH.