There is a titration of 5.00 mL .010 M NaOH with .005 M HCl. Indicate the pH of the initial and final solutions and the pH at the stoichiometric point. What volume of HCl has been added at the stoichiometric point
and the halfway point of the titration?
I have no idea how to solve this problem. I know that the stoichiometric point is always at 7 for strong acid-strong base reactions, but other than that I have no idea where to start.
The stoichiometry point is when moles acid = moles base.
moles base = M x L = ?
moles acid = moles base
moles acid = M x L. You know moles and M, solve for L and convert to mL.
To solve this problem, we need to use the concept of titration and the neutralization reaction between NaOH and HCl.
Let's start by understanding the process of titration in this case. In a titration, a known volume and concentration of one solution (in this case, HCl) is added gradually to another solution (in this case, NaOH) until the reaction between the two is complete. The stoichiometric point in a titration is when the moles of acid and base are in equal amounts, indicating complete reaction.
Given information:
- Volume of NaOH solution = 5.00 mL
- Concentration of NaOH solution = 0.010 M (moles per liter)
- Concentration of HCl solution = 0.005 M
To find the pH of the initial and final solutions, we need to determine the amount of excess HCl or NaOH in the solution after reaction at the stoichiometric point.
1. Initial solution (pure NaOH):
Since NaOH is a strong base, it dissociates completely in water to form OH- ions. Therefore, the initial pH of NaOH is calculated using the formula:
pOH = -log[OH-] = -log(0.010 M) = 2
Recall that the pH and pOH are related by the equation: pH + pOH = 14. Thus, the initial pH of NaOH solution is:
pH = 14 - pOH = 14 - 2 = 12.
2. Stoichiometric point:
At the stoichiometric point, the moles of HCl and NaOH used are equal. This means that all the NaOH has reacted. To calculate the volume of HCl at the stoichiometric point, we need to set up an equation using the mole-to-mole ratio of the neutralization reaction:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
The balanced equation shows that 1 mole of HCl reacts with 1 mole of NaOH to produce 1 mole of NaCl and 1 mole of water. Since the concentration of HCl is 0.005 M and the volume of NaOH is 5.00 mL (0.005 L):
Moles of HCl = Molarity x Volume = 0.005 M x 0.005 L = 0.000025 moles
Therefore, at the stoichiometric point, 0.000025 moles of HCl have been added.
3. Final solution (excess HCl):
Since the reaction between NaOH and HCl is complete at the stoichiometric point, any remaining HCl will be in excess. So, to calculate the pH of the final solution, we need to determine the concentration and volume of the remaining unreacted HCl.
To do that, we subtract the moles of NaOH used in the reaction (0.000025 moles) from the initial moles of NaOH (0.010 M x 0.005 L = 0.00005 moles).
Moles of unreacted NaOH = Initial moles - Stoichiometric moles
= 0.00005 moles - 0.000025 moles = 0.000025 moles
Volume of unreacted NaOH = Stoichiometric moles / Concentration of NaOH
= 0.000025 moles / 0.010 M = 0.0025 L = 2.50 mL
Now we need to calculate the concentration of excess HCl at the stoichiometric point by subtracting the volume of unreacted NaOH (2.50 mL) from the total volume of HCl added:
Volume of HCl at stoichiometric point = Total Volume - Volume of unreacted NaOH
= 5.00 mL - 2.50 mL = 2.50 mL
4. pH at the stoichiometric point:
At the stoichiometric point, the neutralization reaction is complete and only excess HCl remains. Since HCl is a strong acid, it dissociates completely in water to form H+ ions. Therefore, the pH at the stoichiometric point is determined using the HCl concentration:
pH = -log[H+]
= -log(0.005 M)
= 2.3
5. Halfway point of the titration:
The halfway point of the titration occurs when half of the moles of HCl required to react with NaOH have been added. Since the moles of HCl and NaOH react in a 1:1 ratio, the halfway point corresponds to the stoichiometric moles divided by 2:
Moles at halfway point = Stoichiometric moles / 2
= 0.000025 moles / 2
= 0.0000125 moles
To determine the volume of HCl at the halfway point, we can use the mole-to-mole ratio and the concentration of HCl:
Volume of HCl at halfway point = Moles at halfway point / Concentration of HCl
= 0.0000125 moles / 0.005 M
= 0.0025 L = 2.50 mL
Lastly, to find the pH at the halfway point, we use the remaining unreacted HCl moles (0.0000125 moles) and divide it by the total volume of the solution at the halfway point (5.00 mL):
Concentration of HCl = Moles at halfway point / Volume at halfway point
= 0.0000125 moles / 0.005 L
= 0.0025 M
pH at the halfway point = -log(concentration of HCl)
= -log(0.0025 M)
= 2.6
To summarize:
- The initial pH of the NaOH solution is 12.
- The pH at the stoichiometric point is 2.3, indicating the presence of excess HCl.
- The pH at the halfway point is 2.6.
- The volume of HCl at the stoichiometric point is 2.50 mL.
- The volume of HCl at the halfway point is also 2.50 mL.