1. 25mL of 0.1M HCl is titrated with 0.1M NaOH. What is the pH when 30mL of NaOH have been added?
2. 25mL of 0.1M HCl is titrated with 0.1M NaOH. What is the pH after 15mL of NaOH have been added?
I don't know how to make calculation I assumed that for number 1 when you have added 25 mL of NaOH the pH would be neutral (7) but after you add the rest of the 5mL it would be greater than 7. However, we're suppose to give a specific pH value.
#2 I guessed that the pH would be very low. But how do make the calculations?
You're thinking is on the right track for #1. #2 is the same kind of problem.
For both, calculate moles of HCl initially, (moles = M x L), then calculate moles NaOH added. In the first case, the excess hydroxide will be in moles, divide that by the total volume and that give you molarity of OH^-. Calculate pOH and pH from that.
For #2 it's just the reverse. moles HCl initially - moles OH added leaves an excess of HCl unreacted, moles HCl remaining divided by total volume = molarity and from there you get pH. Post back if you get stuck.
I'm sooo stuck.
To calculate the pH in a titration, you need to consider the stoichiometry of the reaction between the acid (HCl) and the base (NaOH). In this case, the stoichiometry is 1:1, meaning that one mole of HCl reacts with one mole of NaOH.
1. To determine the pH when 30 mL of 0.1M NaOH is added to 25 mL of 0.1M HCl, you need to find the amount of NaOH that reacts with the HCl.
First, calculate the moles of NaOH added:
Moles of NaOH = volume of NaOH (in L) × concentration of NaOH
Moles of NaOH = (30 mL / 1000 mL/L) × 0.1 mol/L = 0.003 moles
Since the stoichiometry is 1:1, the moles of NaOH that reacts with HCl is also 0.003 moles.
Now, you need to determine the moles of HCl that are left after the reaction. Since the initial amount of HCl is 0.025 L (25 mL / 1000 mL/L) and the stoichiometry is 1:1, the moles of HCl left are also 0.003 moles.
The pH of the solution depends on the concentration of H+ ions. The moles of H+ ions in solution are equal to the moles of HCl left, which is 0.003 moles.
To determine the concentration of H+ ions, divide the moles by the total volume of the solution:
Concentration of H+ ions = moles of H+ ions / total volume of solution (in L)
Concentration of H+ ions = 0.003 moles / (25 mL + 30 mL) / 1000 mL/L = 0.04 mol/L
Now we can calculate the pH using the formula:
pH = -log[H+]
pH = -log(0.04) ≈ 1.4
Therefore, the pH when 30 mL of NaOH have been added is approximately 1.4.
2. Similarly, to calculate the pH after adding 15mL of NaOH, follow the same steps as above. The only difference is the amount of NaOH added.
Calculate the moles of NaOH added:
Moles of NaOH = (15 mL / 1000 mL/L) × 0.1 mol/L = 0.0015 moles
Since the stoichiometry is 1:1, the moles of NaOH that reacts with HCl is also 0.0015 moles.
Calculate the moles of HCl left:
Moles of HCl left = 0.025 moles (initial amount) - 0.0015 moles (reacted with NaOH) = 0.0235 moles
Calculate the concentration of H+ ions:
Concentration of H+ ions = 0.0235 moles / (25 mL + 15 mL) / 1000 mL/L = 0.094 mol/L
Calculate the pH:
pH = -log(0.094) ≈ 1.0
Therefore, the pH after adding 15 mL of NaOH is approximately 1.0.
Keep in mind that these calculations assume ideal behavior of both the acid and the base, and real experimental conditions may cause some deviation from these results.