A student begins with 25 mL of a 0.434 M solution of HI and slowly adds a solution of 0.365 M NaOH.
1. What is the pH after 15.00 ml of NaOH solution has been added?
2. What is the pH after 40 ml of NaOH solution has been added?
You must know where the equivalence point is in order to know where you are on the titration curve.
mL acid x M acid = mL base x M base.
25*0.434 = mL x 0.365
mL base = estimated 30 mL so 15 mL is BEFORE the eq point and 40 mL is AFTER the eq point.
A. mols HI = M x L = ?
mols NaOH = M x L (0.015 x 0.365) = ?
Subtract, giving you mols HI, divide by total L (15 mL + 25 mL) to give M and convert to pH.
B. mols HI = M x L
mols NaOH= M x L
Subtract giving mols NaOH and divide by total L for M, then convert to pOH and pH.
1.mole of NaOH added = cv = 0.365x15e-3 = 5.48e-3 moles.
Reaction;
HI + NaOH --> NaI + H2O
pH = -log[H+]
the [H+] is the concentration of the acid as 15ml NaOH has been added.
the equation states that, 1 mole of NaOH neutralize 1 mole of HI. So, after adding 15ml, we introduce 5.48e-3 moles of NaOH in the solution. So, this mole should also neutralize 5.48e-3moles of HI.
concentration is moles/volume. The volume of acid is 25mL. with that, use -log[H+].
2. again, find the mole for 40ml of NaOH to find the mole of HI been neutralized. then use that mole to calculate the concentration..
hope that helpss..
To find the pH after adding NaOH solution, we need to understand the reaction of HI (hydroiodic acid) with NaOH (sodium hydroxide) and the concept of acid-base titration.
1. To determine the pH after adding 15.00 ml of NaOH solution:
- Start by calculating the moles of HI present in the initial solution.
Moles of HI = initial volume (in L) * initial concentration (in M)
Initial volume = 25 mL = 0.025 L
Initial concentration = 0.434 M
Moles of HI = 0.025 L * 0.434 M = 0.01085 mol
- Next, calculate the moles of NaOH added.
Moles of NaOH = volume added (in L) * concentration of NaOH (in M)
Volume added = 15.00 mL = 0.015 L
Concentration of NaOH = 0.365 M
Moles of NaOH = 0.015 L * 0.365 M = 0.0055 mol
- Now, determine the limiting reactant. Since HI and NaOH react in a 1:1 ratio, the limiting reactant is the one with fewer moles. In this case, HI is the limiting reactant because it has fewer moles (0.01085 mol vs. 0.0055 mol for NaOH).
- Calculate the moles of excess HI (HI remaining after the reaction).
Moles of excess HI = initial moles of HI - moles of NaOH reacted
Moles of excess HI = 0.01085 mol - 0.0055 mol = 0.00535 mol
- Calculate the new volume of the solution.
New volume = initial volume + volume added
New volume = 0.025 L + 0.015 L = 0.04 L
- Finally, calculate the concentration of HI in the new volume.
Concentration of HI = moles of excess HI / new volume
Concentration of HI = 0.00535 mol / 0.04 L = 0.13375 M
- The pH of the solution can be calculated by using the equation:
pH = -log[H+]
[H+] = concentration of HI (since it's a strong acid)
pH = -log(0.13375) = 0.874
Therefore, the pH after adding 15.00 ml of NaOH solution is approximately 0.874.
2. To determine the pH after adding 40 ml of NaOH solution, follow the same steps as above, but adjust the volume and concentration calculations.
- Initial volume = 25 mL = 0.025 L
- Initial concentration = 0.434 M
- Volume added = 40.00 mL = 0.04 L
- Concentration of NaOH = 0.365 M
- Moles of HI = 0.025 L * 0.434 M = 0.01085 mol
- Moles of NaOH = 0.04 L * 0.365 M = 0.0146 mol (since more NaOH is added this time)
- Moles of excess HI = 0.01085 mol - 0.0146 mol = -0.00375 mol (since NaOH is in excess)
- New volume = initial volume + volume added = 0.025 L + 0.04 L = 0.065 L
- Since there are no excess HI moles remaining, the concentration of HI is 0 M and the pH is not defined (pH value is negative infinity).
Therefore, the pH after adding 40 ml of NaOH solution is undefined.