When 50.0ml of 1.00M HCl is titrated with 1.00M NaOH, the pH increases. calculate the difference in pH of the system when you add 49.99ml NaOH and 50.01ml of NaOH. Please work out detailed steps.

millimols HCl = 50.00 x 1.00M = ?

mmols NaOH = 49.99 x 1.00 M = ?
Subtract to obtain mmols H^+ left, mmols H^+/(50.00+49.99) and convert to pH.

Do the same for 50.00 and 50.01, subtract to find excess OH^- mmols, then
OH^- = mmols OH/total mL and convert to pOH and pH.

No

To calculate the difference in pH of the system when you add 49.99 mL NaOH and 50.01 mL of NaOH, we need to consider the reaction between HCl (acid) and NaOH (base). First, let's determine the moles of HCl and NaOH in the reaction.

Step 1: Calculate the moles of HCl.
Given:
- Volume of HCl (acid) = 50.0 mL
- Concentration of HCl = 1.00 M

Moles of HCl = Concentration × Volume
Moles of HCl = 1.00 M × 50.0 mL
Moles of HCl = 0.050 mol

Step 2: Calculate the moles of NaOH for each addition.
Given:
- Volume of NaOH (base) = 49.99 mL and 50.01 mL
- Concentration of NaOH = 1.00 M

Moles of NaOH = Concentration × Volume
Moles of NaOH = 1.00 M × 49.99 mL (for the first addition)
Moles of NaOH = 0.04999 mol

Moles of NaOH = 1.00 M × 50.01 mL (for the second addition)
Moles of NaOH = 0.05001 mol

Step 3: Find the limiting reactant.
Since HCl is in excess, NaOH is the limiting reactant in this reaction. We only need to consider the moles of NaOH.

Step 4: Calculate the moles of HCl consumed for each addition.
Moles of HCl consumed = Moles of NaOH
Moles of HCl consumed (for the first addition) = 0.04999 mol
Moles of HCl consumed (for the second addition) = 0.05001 mol

Step 5: Calculate the remaining moles of HCl.
Remaining moles of HCl = Initial moles of HCl - Moles of HCl consumed

For the first addition:
Remaining moles of HCl (for the first addition) = 0.050 mol - 0.04999 mol
Remaining moles of HCl (for the first addition) = 0.00001 mol

For the second addition:
Remaining moles of HCl (for the second addition) = 0.050 mol - 0.05001 mol
Remaining moles of HCl (for the second addition) = -0.00001 mol

It is important to note that the negative value for the second addition was obtained due to the assumption that excess NaOH was added. In reality, this is not possible, and it indicates an error in the calculations or measurement precision.

Step 6: Calculate the pH for each addition.
To calculate the pH, we need to determine the concentration of HCl remaining in each case. We can use the remaining moles of HCl and the total volume of the solution.

For the first addition:
Volume of the solution = Volume of HCl + Volume of NaOH (base)
Volume of the solution = 50.0 mL + 49.99 mL
Volume of the solution = 99.99 mL

Concentration of HCl remaining = Remaining moles of HCl / Volume of the solution
Concentration of HCl remaining (for the first addition) = 0.00001 mol / 99.99 mL

For the second addition:
Volume of the solution = Volume of HCl + Volume of NaOH (base)
Volume of the solution = 50.0 mL + 50.01 mL
Volume of the solution = 100.01 mL

Concentration of HCl remaining = Remaining moles of HCl / Volume of the solution
Concentration of HCl remaining (for the second addition) = -0.00001 mol / 100.01 mL

Again, note that the negative value for the second addition is not possible in a real scenario.

Step 7: Convert the concentration of HCl remaining to pH.
To convert concentration to pH, we can calculate the negative logarithm (base 10) of the concentration.

pH = -log10(Concentration of HCl remaining)

For the first addition:
pH (for the first addition) = -log10(Concentration of HCl remaining (for the first addition))

For the second addition:
pH (for the second addition) = -log10(Concentration of HCl remaining (for the second addition))

By following these steps, you can calculate the difference in pH of the system when you add 49.99 mL NaOH and 50.01 mL NaOH. Remember to consider the limitations and not to rely on negative concentrations that may arise due to assumptions or calculation errors.