For the titration of 25.00 mL of 0.150 M HCl with 0.250 M NaoH, calculate (a) the initial pH; (b) the pH when neutralization is 50% complete; (c) the pH when neutralization is 100% complete; and (d) the pH when 1.00 mL of NaOH is added beyond the equivalence point.
Here is what you need to do. Determine what you have in solution at each of the four points. To do that, determine moles HCl initially, moles NaOH added, subtract to see which is in excess and go from there. Post your work if you get stuck.
I determined the number of millimoles of HCl initially to be 3.75.
How do I determine how much moles of NaOH are added at 50% completion for example because the volume isn't given.
The reaction is:
HCl + NaOH --> NaCl + H2O
For 100% neutralization, moles NaOH needed is 3.75.
For 50% neutralization, you use half of 3.75 millimoles NaOH. At that point, 50% of the HCl is still there.
[H+] = conc. of HCl = (moles HCl)/(Total volume)*
*Note: Total volume (L) = 0.02500L+(0.001875mol/(0.250mol/L)
Ok Thanks a lot guys.
To answer these questions, we need to understand the titration process and how to calculate the pH at different stages. Let's go through each part step by step.
(a) To calculate the initial pH of the solution, we need to determine the concentration of H+ ions in the HCl solution. Since HCl is a strong acid, it completely ionizes in water to form H+ and Cl- ions. Therefore, the concentration of H+ ions is equal to the concentration of HCl, which is 0.150 M in this case.
Taking the -log of the H+ concentration, we can calculate the initial pH using the formula:
pH = -log[H+]
So, pH = -log(0.150) = 0.823.
Therefore, the initial pH of the solution is approximately 0.823.
(b) To determine the pH when neutralization is 50% complete, we need to find the point where half of the HCl is neutralized by NaOH. At this point, the moles of HCl neutralized is equal to half of the initial moles of HCl.
First, let's calculate the initial moles of HCl:
Moles = concentration × volume
Moles = 0.150 M × 0.02500 L = 0.00375 moles
Since the ratio between HCl and NaOH is 1:1, half of the moles of HCl will be neutralized when 0.001875 moles of NaOH is added.
Now, let's calculate the remaining moles of HCl:
Remaining moles = initial moles - neutralized moles
Remaining moles = 0.00375 moles - 0.001875 moles = 0.001875 moles
Next, we need to find the excess moles of NaOH present after neutralization is 50% complete. Since half of the initial moles of HCl is neutralized, the NaOH will be in excess by the same amount.
Excess moles of NaOH = neutralized moles of HCl = 0.001875 moles
To find the remaining concentration of H+ ions, divide the remaining moles of HCl by the total volume of the solution:
Concentration = remaining moles / total volume
Concentration = 0.001875 moles / (25.00 mL + 1.00 mL) = 0.001783 M
Now we can calculate the pH using the same formula as before:
pH = -log[0.001783] ≈ 2.749.
Therefore, the pH when neutralization is 50% complete is approximately 2.749.
(c) When neutralization is 100% complete, all the HCl will be completely neutralized by the NaOH. In this case, we have a solution of NaCl (a salt). NaCl is a neutral salt, so it does not produce any H+ or OH- ions when dissolved in water. Therefore, the pH of the solution will be equal to 7, which is neutral.
Hence, the pH when neutralization is 100% complete is 7.
(d) When 1.00 mL of NaOH is added beyond the equivalence point, we have excess OH- ions in the solution. Now, we need to calculate the moles of OH- ions present in 1.00 mL of 0.250 M NaOH:
Moles = concentration × volume
Moles = 0.250 M × 0.00100 L = 0.00025 moles
Since we have an excess of OH- ions, we need to consider the remaining concentration of the OH- ions. To do this, we subtract the moles of OH- ions from the excess moles of NaOH:
Remaining moles of OH- = Excess moles of NaOH - Moles of OH-
Remaining moles of OH- = 0.001875 moles - 0.00025 moles = 0.001625 moles
To calculate the remaining concentration of OH- ions, divide the remaining moles by the total volume of the solution:
Concentration = remaining moles / total volume
Concentration = 0.001625 moles / (25.00 mL + 1.00 mL) = 0.001544 M
Now, we need to calculate the pOH using the formula:
pOH = -log[OH-]
pOH = -log[0.001544] ≈ 2.811
Finally, to find the pH, we can use the relationship between pH and pOH:
pH + pOH = 14
pH = 14 - pOH
pH = 14 - 2.811 ≈ 11.189
Therefore, the pH when 1.00 mL of NaOH is added beyond the equivalence point is approximately 11.189.