24.00 mL of a 0.25 M NaOH solution is titrated with 0.10M HCl. What is the pH of the solution after 24.00 mL of the HCl has been added?
Why did the pH and the titration solution go to couples therapy? Because they needed to find some common ground!
To find the pH of the solution after adding 24.00 mL of 0.10 M HCl to 24.00 mL of 0.25 M NaOH, we need to determine the moles of HCl and NaOH, and then calculate the excess or remaining concentration of the resulting solution.
Step 1: Calculate the moles of HCl added.
Moles of HCl = Volume of HCl (in liters) × Concentration of HCl
Moles of HCl = 0.024 L × 0.10 mol/L
Moles of HCl = 0.0024 moles
Step 2: Calculate the moles of NaOH in the initial solution.
Moles of NaOH = Volume of NaOH (in liters) × Concentration of NaOH
Moles of NaOH = 0.024 L × 0.25 mol/L
Moles of NaOH = 0.006 moles
Step 3: Determine the limiting reagent.
To identify the limiting reagent, we compare the moles of HCl and NaOH.
We can see that the moles of HCl (0.0024 mol) are lower than the moles of NaOH (0.006 mol). Therefore, HCl is the limiting reagent.
Step 4: Determine the excess moles of NaOH.
Excess moles of NaOH = Moles of NaOH - Moles of HCl
Excess moles of NaOH = 0.006 mol - 0.0024 mol
Excess moles of NaOH = 0.0036 mol
Step 5: Calculate the concentration of the resulting solution.
Total volume of the resulting solution = Volume of HCl + Volume of NaOH
Total volume = 0.024 L + 0.024 L
Total volume = 0.048 L
Concentration of the resulting solution = Excess moles of NaOH / Total volume
Concentration of the resulting solution = 0.0036 mol / 0.048 L
Concentration of the resulting solution = 0.075 M
Step 6: Calculate the pOH of the resulting solution.
pOH = -log10 (Concentration of hydroxide ions)
Since NaOH is a strong base, it dissociates completely in water, producing one hydroxide ion for every one mole of NaOH.
pOH = -log10 (0.075)
pOH = 1.12
Step 7: Calculate the pH of the resulting solution.
pOH + pH = 14
pH = 14 - pOH
pH = 14 - 1.12
pH = 12.88
Therefore, the pH of the solution after adding 24.00 mL of the 0.10 M HCl is approximately 12.88.
To determine the pH of the solution after adding the HCl, we need to calculate the concentration of the remaining OH- ions in the solution.
First, let's determine the number of moles of NaOH in the initial solution:
Moles of NaOH = volume of NaOH solution (in liters) * concentration of NaOH (in moles/L)
= 0.024 L * 0.25 mol/L
= 0.006 mol
Since the balanced chemical equation for the reaction between NaOH and HCl is:
NaOH + HCl → NaCl + H2O
The moles of NaOH will react with an equal number of moles of HCl. Therefore, 0.006 moles of HCl will react with 0.006 moles of NaOH.
Now, let's calculate the remaining moles of NaOH after the reaction:
Remaining moles of NaOH = Initial moles of NaOH - Moles of HCl reacted
= 0.006 mol - 0.006 mol
= 0 mol
Since all the NaOH has reacted with HCl, the concentration of OH- ions is now 0. Therefore, the pH of the solution is determined by the concentration of H+ ions.
The reaction between HCl and water produces H+ ions:
HCl + H2O → H3O+ + Cl-
Since the original concentration of HCl solution is 0.10 mol/L and 0.024 L has been added, the moles of HCl added are:
Moles of HCl added = volume of HCl solution (in liters) * concentration of HCl (in moles/L)
= 0.024 L * 0.10 mol/L
= 0.0024 mol
The total moles of HCl in the solution are the moles of HCl initially present plus the moles of HCl added:
Total moles of HCl = Moles of HCl initially present + Moles of HCl added
= 0 mol + 0.0024 mol
= 0.0024 mol
The concentration of H+ ions can be calculated by dividing the total moles of HCl by the total volume of the solution:
Concentration of H+ ions = Total moles of HCl / Total volume of solution (in liters)
= 0.0024 mol / 0.048 L
= 0.05 mol/L
To convert the concentration of H+ ions to pH, we use the equation:
pH = -log10(H+)
pH = -log10(0.05)
pH ≈ 1.30
Therefore, the pH of the solution after adding 24.00 mL of 0.10 M HCl is approximately 1.30.
millimoles NaOH = mL x M = 24.00 mL x 0.25 M = 6.00
millimoles HCl = 24.00 mL x 0.10 M = 2.40
total volume = 48.00 mL
.................................NaOH + HCl ==>NaCl + H2O
initial.........................6.00.........0............0.........0
added.....................................2.40............................
change.................... -2.40.....-2.40.........+2.40....+2.40
equilibrium.................3.60.........0..............2.40.......2.40
The NaCl contributes nothing to the pH of the final solution. The pH is determined by the excess of NaOH present. (NaOH) = millimoles/mL = 3.60/48.00 = ? M = (OH^-)
pOH = -log (OH^-). Then
pH + pOH = pKw = 14.00. You know pOH from above and you know pKw = 14; substitute and solve for pH. Post your work if you get stuck.