Consider the titration of 100.0 mL of 0.110 M NaOH with 1.00 M HBr. Find the pH at the following volumes of acid added and make a graph of pH versus Va:Va = 0,2,5,8,10,10.1,11,11.1, and 12mL. Compute Ve, first.
Find mols NaOH, subtract mols HBr.
If mols NaOH are greater then pOH = -log(OH^-) and convert to pH by
pH + pOH = pKw = 14.
If mols HBr are in excess, pH = -log(H^+).
Those three steps will do all of the points except for the equivalence point. That one is done by the hydrolysis of the salt, NaBr, which is 7 at the equivalence point.
To determine the pH at each volume of acid added, we can use the concept of stoichiometry in the acid-base reaction between NaOH and HBr. The balanced equation for the reaction is:
NaOH + HBr -> NaBr + H2O
Given that the initial volume of NaOH is 100.0 mL and the concentration is 0.110 M, we can calculate the number of moles of NaOH:
moles of NaOH = volume of NaOH (L) × concentration of NaOH (M)
= 0.100 L × 0.110 mol/L
= 0.011 mol
Since the stoichiometry of NaOH and HBr is 1:1, the number of moles of HBr required to reach the equivalence point (Ve) would also be 0.011 mol.
To find the volume of HBr required to reach the equivalence point, we can use the equation:
moles of HBr = volume of HBr (L) × concentration of HBr (M)
Plugging in the values:
0.011 mol = volume of HBr (L) × 1.00 mol/L
Therefore, the volume of HBr required to reach the equivalence point (Ve) is 0.011 L or 11.0 mL.
Now, let's calculate the pH at each of the specified volumes of HBr added and create a graph of pH versus Va:
1. For Va = 0 mL (before any acid is added)
- Moles of NaOH remaining = moles of NaOH initially = 0.011 mol
- Volume of solution = 100 mL + 0 mL = 100 mL = 0.100 L
- Concentration of NaOH remaining = moles of NaOH remaining / volume of solution
= 0.011 mol / 0.100 L
= 0.110 M (unchanged)
Since NaOH is a strong base, it dissociates completely in water. Thus, [OH-] = 0.110 M.
Using the equation for pH: pH = -log[H+]
[H+] = Kw / [OH-] = 1.00 × 10^-14 / 0.110 = 9.09 × 10^-14 M
pH = -log(9.09 × 10^-14) ≈ 13.04
2. For Va = 2 mL
At this point, all the NaOH has not yet reacted, and we need to find the concentration of OH- ions remaining.
moles of NaOH remaining = 0.011 mol - (2 mL / 1000 mL/L) × 1.00 M = 0.009 mol
Volume of solution = 100 mL + 2 mL = 102 mL = 0.102 L
Concentration of NaOH remaining = 0.009 mol / 0.102 L ≈ 0.088 M
[OH-] = 0.088 M
[H+] ≈ Kw / [OH-] ≈ 1.00 × 10^-14 / 0.088 ≈ 1.14 × 10^-13 M
pH ≈ -log(1.14 × 10^-13) ≈ 12.94
3. For Va = 5 mL, 8 mL, 10 mL, 10.1 mL, 11 mL, 11.1 mL
We can follow the same procedure as above to find the concentration of OH- ions remaining and calculate the pH at each point using the equation: pH = -log[H+].
4. For Va = 12 mL (Ve)
The volume of HBr added has reached the equivalence point, which means all the NaOH has reacted completely with HBr. Therefore, the pH at the equivalence point (Ve) would be determined by the salt formed in the reaction (NaBr).
Notice that at the equivalence point, NaBr is a neutral salt. Therefore, the pH at the equivalence point (Ve) is approximately 7.
Now, we can plot the graph of pH versus Va using the values of Va and pH determined at each point.
However, please note that the pH values at points other than the equivalence point (Ve) will depend on the ranges provided. If you need pH values for specific Va values within the given range, please provide the specific values, and we can calculate and plot them accordingly.
To find the pH at each volume of acid added and create a graph of pH versus Va, we need to understand the process of titration and how to calculate the pH.
1. Understand the titration process:
Titration is a technique used to determine the concentration of a solution by gradually adding a solution with a known concentration (titrant) to a solution of unknown concentration (analyte). In this case, we are titrating NaOH (analyte) with HBr (titrant).
2. Calculate the equivalence point (Ve):
The equivalence point, Ve, is the volume of titrant required to completely react with the analyte. In this case, since NaOH and HBr react in a 1:1 ratio, the equivalence point is when the moles of NaOH are equal to the moles of HBr.
To find Ve, we can use the equation:
M1V1 = M2V2
Where M1 and V1 are the initial molarity and volume of NaOH, and M2 and V2 are the molarity and volume of HBr at the equivalence point.
Given:
M1 = 0.110 M (NaOH)
V1 = 100.0 mL
M2 = 1.00 M (HBr)
Using the equation, we can calculate Ve:
(0.110 M)(100.0 mL) = (1.00 M)(Ve)
Ve = (0.110 M)(100.0 mL) / (1.00 M)
Ve = 11.0 mL
Therefore, the equivalence point volume (Ve) is 11.0 mL.
3. Calculate the pH at each volume of acid added:
To calculate the pH of the solution at different volumes of acid added, we need to consider the reaction between NaOH and HBr. At the start of the titration, before any HBr has been added, the solution consists of 0.110 M NaOH. As HBr is added, it reacts with NaOH to form water (H2O) and a salt (NaBr).
The balanced chemical equation for the reaction is:
NaOH + HBr -> NaBr + H2O
The reaction between a strong base (NaOH) and a strong acid (HBr) results in a neutral solution (pH = 7) at the equivalence point (Ve). However, before reaching the equivalence point, the pH will be influenced by the excess NaOH or HBr in the solution.
To calculate the pH, we need to determine the concentration of the remaining base or acid at each volume (Va) of acid added.
Given the initial concentration and volume of NaOH, we can use the following equation to calculate the concentration of NaOH remaining at each volume (Va) of HBr added:
C1V1 = C2V2
Where C1 and V1 are the initial concentration and volume of NaOH, and C2 and V2 are the concentration and volume of NaOH remaining after adding Va mL of HBr.
Given:
C1 = 0.110 M (NaOH)
V1 = 100.0 mL
Now, let's calculate the remaining concentration of NaOH at each volume of HBr added (Va).
For Va = 0 mL:
C2 = C1 (initial concentration)
pH = -log(C2)
For Va = 2 mL:
Calculate the new volume of NaOH:
V2 = V1 - Va
Calculate the new concentration of NaOH:
C2 = C1 * (V1 / V2)
pH = -log(C2)
Repeat the above steps for Va = 5 mL, 8 mL, 10 mL, 10.1 mL, 11 mL, 11.1 mL, and 12 mL.
4. Create a graph of pH versus Va:
Once you have calculated the pH at each volume of acid added, plot the values on a graph with Va (volume of acid added) on the x-axis and pH on the y-axis. Connect the plotted points to create the graph.
Note: The pH values will likely change abruptly around the equivalence point (Ve) due to the sudden shift from acidic to basic or basic to acidic conditions.