Hi! I am very stuck with these problems. I have tried to solve for each of these values, but I keep getting the answers wrong.
Q:
For each of the following strong base solutions determine OH-, H3O+, pH, and pOH:
8.66×10−3 M LiOH
1.15×10−2 M Ba(OH)2
2.2×10−4 M KOH
4.9×10−4 M Ca(OH)2
To determine the values of OH-, H3O+, pH, and pOH for each of the given strong base solutions, you need to understand the concepts of pH and pOH and how they are related to the hydroxide ion concentration (OH-) and hydronium ion concentration (H3O+).
To start, let's define pH and pOH:
- pH: It is a measure of the concentration of H3O+ ions in a solution. It is calculated using the formula: pH = -log[H3O+].
- pOH: It is a measure of the concentration of OH- ions in a solution. It is calculated using the formula: pOH = -log[OH-].
Furthermore, the pH and pOH values are related by the equation:
pH + pOH = 14
Now, let's solve each problem step by step:
1. 8.66×10−3 M LiOH:
Since LiOH is a strong base, it completely dissociates into Li+ ions and OH- ions in solution. Therefore, the concentration of OH- ions is equal to the concentration of LiOH.
OH- concentration = 8.66×10−3 M
To find pOH, you can use the formula pOH = -log[OH-]. Plug in the OH- concentration you calculated:
pOH = -log(8.66×10−3) ≈ 2.06
Since pH + pOH = 14, you can calculate pH:
pH = 14 - pOH = 14 - 2.06 ≈ 11.94
2. 1.15×10−2 M Ba(OH)2:
Again, since Ba(OH)2 is a strong base, it completely dissociates into Ba2+ ions and OH- ions in solution. Therefore, the concentration of OH- ions is twice the concentration of Ba(OH)2.
OH- concentration = 2 × 1.15×10−2 M = 2.3×10−2 M
Now, using the same formulas as before:
pOH = -log(2.3×10−2) ≈ 1.64
pH = 14 - pOH = 14 - 1.64 ≈ 12.36
3. 2.2×10−4 M KOH:
Once again, KOH is a strong base, so it fully dissociates into K+ ions and OH- ions in solution.
OH- concentration = 2.2×10−4 M
Using the formulas:
pOH = -log(2.2×10−4) ≈ 3.66
pH = 14 - pOH = 14 - 3.66 ≈ 10.34
4. 4.9×10−4 M Ca(OH)2:
As before, Ca(OH)2 is a strong base and will completely dissociate into Ca2+ ions and twice the concentration of OH- ions.
OH- concentration = 2 × 4.9×10−4 M = 9.8×10−4 M
Using the formulas:
pOH = -log(9.8×10−4) ≈ 3.01
pH = 14 - pOH = 14 - 3.01 ≈ 10.99
In summary:
8.66×10−3 M LiOH: OH- concentration = 8.66×10−3 M, pOH ≈ 2.06, pH ≈ 11.94
1.15×10−2 M Ba(OH)2: OH- concentration = 2.3×10−2 M, pOH ≈ 1.64, pH ≈ 12.36
2.2×10−4 M KOH: OH- concentration = 2.2×10−4 M, pOH ≈ 3.66, pH ≈ 10.34
4.9×10−4 M Ca(OH)2: OH- concentration = 9.8×10−4 M, pOH ≈ 3.01, pH ≈ 10.99
Remember to double-check your calculations, and pay attention to significant figures to ensure accurate results.