Calculate the pH after 0.15 mol of NaOH is added to 1.00 L of the solution that is 0.50 M HF and 1.01 M KF
To calculate the pH after adding NaOH to the solution, we need to determine the concentration of the remaining HF (hydrofluoric acid) and calculate the concentration of OH^− ions formed from the reaction between NaOH and HF.
First, let's write the balanced chemical equation for the reaction between NaOH and HF:
NaOH + HF → NaF + H2O
From this equation, we can see that 1 mole of NaOH reacts with 1 mole of HF to produce 1 mole of NaF and 1 mole of water.
Given that 0.15 mol of NaOH is added to 1.00 L of the solution containing 0.50 M HF and 1.01 M KF, we need to determine the number of moles of HF that initially exist in the solution:
moles of HF = concentration × volume
moles of HF = 0.50 M × 1.00 L = 0.50 mol
Since the balanced equation tells us the stoichiometry between NaOH and HF is 1:1, we know that 0.15 mol of HF is reacting with 0.15 mol of NaOH. Therefore, we subtract the 0.15 mol from the initial 0.50 mol to find the remaining moles of HF:
remaining moles of HF = initial moles of HF - moles of HF reacted
remaining moles of HF = 0.50 mol - 0.15 mol = 0.35 mol
Next, we need to determine the concentration of HF in the solution after the reaction:
final concentration of HF = remaining moles of HF / volume of solution
final concentration of HF = 0.35 mol / 1.00 L = 0.35 M
Now, let's consider the reaction between NaOH and HF, which produces NaF and water. This reaction generates OH^− ions. Since 1 mole of NaOH reacts with 1 mole of HF, we will have 0.15 mol of OH^− ions as a result.
Concentration of OH^− ions = moles of OH^− ions / volume of solution
Concentration of OH^− ions = 0.15 mol / 1.00 L = 0.15 M
To calculate the pH, we need to determine the pOH first:
pOH = -log[OH^−]
pOH = -log(0.15) = 0.823
Finally, we can calculate the pH using the equation:
pH = 14 - pOH
pH = 14 - 0.823 = 13.177
Therefore, the pH after adding 0.15 mol of NaOH to 1.00 L of the solution containing 0.50 M HF and 1.01 M KF is approximately 13.177.