A 25 mL sample of 0.29 M LiOH analyte was titrated with 0.75 M HI at 25 degree Celsius. 1)Calculate the initial pH before any titrant was added. 2) Then calculate the pH of the solution after 5 mL of titrant was added.
At the beginning you have 0.29 M LiOH. That ionizes 100% so (OH^-) = 0.29M
pOH = -log(OH^-); solve for pOH,then
pOH + pH = pKw = 14. YOu know pKw and pOH solve for pH.
To start we have 25 mL x 0.29 M = approx 7 millimols LiOH but you need a better answer on this as well as all of the other calculations that follow.
For pH after the addition of 5.0 mL of 0.75M that is 3.75 mmols HI added.
..............LiOH + HI ==> H2O + LiI
I.............7.....0......0.....0
add................3.75
C...........-3.75.-3.75.....3.75..3.75
E......approx3.25..0........3.75..3.75
M LiOH remaining is mmols/mL = approx 3.25/(25+5) = ?
Then convert to pOH followed by convert to pH.
To calculate the initial pH before any titrant was added, we need to determine the concentration of hydroxide ions (OH-) in the LiOH solution.
1) Calculate the initial pH:
First, we need to convert the volume of the LiOH solution to liters:
25 mL = 25/1000 = 0.025 L
Next, use the formula for calculating the initial concentration of OH- ions:
OH- concentration (M) = moles of OH- / volume (L)
We know the moles of OH- can be calculated as:
moles of OH- = concentration (M) × volume (L)
Given that the LiOH concentration is 0.29 M and the volume is 0.025 L, we can calculate the moles of OH-:
moles of OH- = 0.29 M × 0.025 L = 0.00725 mol
Now, we can calculate the OH- concentration:
OH- concentration (M) = 0.00725 mol / 0.025 L = 0.29 M
Since the concentration of OH- is equal to the concentration of LiOH, the initial pH can be determined by calculating the pOH and then subtracting it from 14.
pOH = -log10(OH- concentration)
pOH = -log10(0.29) ≈ 0.537
pH = 14 - pOH = 14 - 0.537 ≈ 13.463
Therefore, the initial pH before any titrant was added is approximately 13.463.
2) To calculate the pH of the solution after 5 mL of titrant was added, we need to determine the concentration of hydronium ions (H3O+).
Since HI is a strong acid, it will completely dissociate in water, producing hydronium ions (H3O+). The reaction between LiOH and HI can be represented as:
LiOH + HI → LiI + H2O
In this reaction, one mole of LiOH reacts with one mole of HI, which means that the initial moles of OH- will be equal to the moles of hydronium ions (H3O+) produced.
Given that 5 mL of HI was added, we need to calculate the moles of H3O+ and convert it to concentration (M) using the total volume of the solution (25 mL + 5 mL).
First, convert the volume of HI to liters:
5 mL = 5/1000 = 0.005 L
Next, calculate the moles of H3O+:
moles of H3O+ = concentration (M) × volume (L)
moles of H3O+ = 0.75 M × 0.005 L = 0.00375 mol
Now, calculate the concentration of H3O+:
H3O+ concentration (M) = moles of H3O+ / total volume (L)
Total volume = 25 mL + 5 mL = 30 mL = 30/1000 = 0.03 L
H3O+ concentration (M) = 0.00375 mol / 0.03 L ≈ 0.125 M
To find the pH, we can calculate the pOH:
pOH = -log10(H3O+ concentration)
pOH = -log10(0.125) ≈ 0.903
Finally, calculate the pH:
pH = 14 - pOH = 14 - 0.903 ≈ 13.097
Therefore, the pH of the solution after 5 mL of titrant was added is approximately 13.097.