Consider the titration of 100.0 mL of 0.100 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 = 0,6 and 12 mL
HBr + NaOH ==> NaBr + H2O
1. First you need to calculate the volume of HBr required to reach the equivalence
mols NaOH = M x L = 0.1M x 0.1L =- 0.01
mols HBr must be 0.01 to reach the equivalence point.
Then M HBr = mols HBr/L HBr. Substitute to obtain 1M = 0.01/L and L = 0.01/1 = 0.01L or 10.0 mL of the 1.00 M HBr. That tells you that the 6.0 mL HBr is BEFORE the eq. pt. and 12.0 mL HBr is AFTER the eq. pt.
a. at zero HBr.
You have 0.1M NaOH; therefore, (OH^-) = 0.1. Convert that to pH.
b. mols NaOH initially = 0.1M x 0.1L = 0.01.
mols HBr added at 6 mL = 1M x 0.006 L = 0.006 mols
0.01 mol NaOH - 0.006 mol HBr = 0.004 and since M = mols/L then 0.004/(0.106) = 0.0377. That is OH^-, convert to pH.
I will leave c for you. That is after the equivalence point.
To find the pH at different volumes of acid added, we need to understand the concept of titration and the reaction that takes place between NaOH and HBr.
During titration, a measured volume of one solution (the titrant) is added to a known volume of another solution (the analyte) until the reaction between the two is complete. In this case, 0.100 M NaOH is the analyte, and 1.00 M HBr is the titrant.
The reaction between NaOH and HBr can be represented as follows:
HBr + NaOH -> NaBr + H2O
At the start of the titration (Va = 0 mL), no acid has been added yet. Therefore, the concentration of HBr is zero. We can calculate the pH of the solution using the equation for the dissociation of water:
H2O -> H+ + OH-
Since there is no HBr (H+) yet, the only source of H+ ions is the auto-ionization of water. At room temperature, the concentration of H+ and OH- is equal to 1.0 x 10^-7 M. Therefore, the pH at Va = 0 mL is equal to -log[H+] = -log[1.0 x 10^-7] = 7.
When 6 mL of HBr is added (Va = 6 mL), we need to determine the moles of HBr and NaOH that react. The balanced equation shows a 1:1 mole ratio between them. Using the formula for moles (moles = concentration x volume), we can calculate the moles of HBr:
Moles of HBr = (concentration of HBr) x (volume of HBr added)
= (1.00 M) x (6 mL / 1000 mL/L)
= 0.006 moles
Since NaOH and HBr react in a 1:1 ratio, 0.006 moles of NaOH will be consumed. The remaining moles of NaOH can be calculated using the initial concentration and the volume of HBr added:
Moles of NaOH remaining = (initial concentration of NaOH) x (initial volume of NaOH - volume of HBr added)
= (0.100 M) x (100 mL - 6 mL) / 1000 mL/L
= 0.0094 moles
To find the new concentration of NaOH, divide the moles remaining by the new volume:
New concentration of NaOH = (moles of NaOH remaining) / (new volume of NaOH remaining)
= 0.0094 moles / (100 mL - 6 mL) / 1000 mL/L
= 0.112 M
Now, we can calculate the concentration of OH- ions using the new concentration of NaOH:
[OH-] = (new concentration of NaOH) = 0.112 M
Since this concentration of OH- ions is higher than the concentration of H+ ions (1.0 x 10^-7 M), the solution will be basic. To find the pH, we can use the equation:
pOH = -log[OH-]
pH = 14 - pOH
pOH = -log[0.112] ≈ 0.950
pH = 14 - 0.950 ≈ 13.05
Finally, for Va = 12 mL, you can repeat the same calculations to find the new concentration of NaOH and then the pH of the solution.
To make a graph of pH versus Va, plot the pH values corresponding to each volume of acid added on the y-axis against the corresponding volumes of acid added on the x-axis.