The pH of a Ba(OH)2 solution is 10.66. What is the hydroxide concentration? If the solution volume is 125ml, how many grams of Ba(OH)2 have been dissolved?
pH = 10.66
pOH = 14 - 10.66 =3.34
(OH^-) = 10-3.34 = 4.57 x 10^-4 M
Actualy, Ba(OH)2 is rather soluble BUT it does have a Ksp in my books. I assume we are to assume it is soluble.
If (OH^-) = 4.57 x 10^-4 M, then [Ba)OH)2] is 1/2 that or 2.28 x 10^-4 M.
M = mols/L. You know M and L, calculate mols.
Then mol = g/molar mass. You know mol and molar mass, calculate grams.
Check my work carefully.
Oh, we've got a pH of 10.66? That's quite basic, if you ask me! Ba(OH)2 is a strong base, so it fully dissociates in water.
Now, to determine the hydroxide concentration, we'll use a little pH formula trickery. Since pH = -log[H+], we can figure out that [H+] = 10^(-pH). In this case, [H+] = 10^(-10.66), which is approximately 2.65 x 10^(-11) M.
But hold your test tubes, we're not done yet! Since Ba(OH)2 dissociates into two OH- per formula unit, the concentration of hydroxide will be twice that of [H+]. Thus, the hydroxide concentration is approximately 5.3 x 10^(-11) M.
Now, onto the second part of the question! If the solution volume is 125 mL, we need to find the number of moles of Ba(OH)2 dissolved. The formula for calculating moles is moles = concentration × volume (in liters). So, moles = (5.3 x 10^(-11) M) × (0.125 L), which is approximately 6.63 x 10^(-12) moles.
Finally, to find the mass of Ba(OH)2, we need to multiply the number of moles by its molar mass. The molar mass of Ba(OH)2 is 171.34 g/mol (Ba: 137.33 g/mol, O: 16.00 g/mol, H: 1.01 g/mol).
So, grams = (6.63 x 10^(-12) moles) × (171.34 g/mol), which comes out to be approximately 1.13 x 10^(-9) grams.
Voila! Around 1.13 nanograms of Ba(OH)2 have been dissolved. Be careful not to sneeze, or it might disappear into thin air!
To find the hydroxide concentration, we can use the formula for pH:
pH = -log[H+]
Since Ba(OH)2 dissociates into two hydroxide ions (OH-), the concentration of hydroxide ions is twice the concentration of barium hydroxide.
1. Calculate the concentration of hydroxide ions (OH-) using the pH:
pOH = 14 - pH
= 14 - 10.66
= 3.34
[OH-] = 10^-pOH
= 10^-3.34
= 4.642 x 10^-4 M
2. Since the concentration of barium hydroxide (Ba(OH)2) is twice the concentration of hydroxide ions, we multiply the concentration of hydroxide ions by 2.
[Ba(OH)2] = 2 * [OH-]
= 2 * 4.642 x 10^-4 M
= 9.284 x 10^-4 M
3. Convert the concentration from molarity (M) to moles per liter (mol/L).
9.284 x 10^-4 M = 9.284 x 10^-4 mol/L
4. Calculate the moles of barium hydroxide (Ba(OH)2) in the solution using the volume.
moles = concentration * volume
= 9.284 x 10^-4 mol/L * 0.125 L
= 1.16 x 10^-4 mol
5. Finally, calculate the grams of barium hydroxide (Ba(OH)2) dissolved in the solution using the molar mass of Ba(OH)2.
Molar mass of Ba(OH)2 = 137.33 g/mol + 2(16.00 g/mol) + 2(1.01 g/mol)
= 137.33 g/mol + 32.00 g/mol + 2.02 g/mol
= 171.35 g/mol
grams = moles * molar mass
= 1.16 x 10^-4 mol * 171.35 g/mol
= 0.0199 grams (rounded to 4 decimal places)
Therefore, the hydroxide concentration is 9.284 x 10^-4 M, and approximately 0.0199 grams of Ba(OH)2 have been dissolved in the 125 ml solution.
To find the hydroxide concentration in a Ba(OH)2 solution, you can use the equation:
pOH = 14 - pH
First, let's calculate the pOH:
pOH = 14 - 10.66
pOH = 3.34
The pOH of the solution is 3.34.
Next, we can convert the pOH to the hydroxide concentration using the following equation:
pOH = -log[OH-]
Rearranging the equation, we get:
[OH-] = 10^(-pOH)
[OH-] = 10^(-3.34)
Using a scientific calculator, we can find:
[OH-] ≈ 4.28 x 10^(-4) M
The hydroxide concentration is approximately 4.28 x 10^(-4) M.
To calculate the number of moles of Ba(OH)2, we can use the hydroxide concentration, balanced chemical equation, and stoichiometry.
The balanced chemical equation for the dissociation of Ba(OH)2 is:
Ba(OH)2 → Ba2+ + 2OH-
From the equation, we see that one mole of Ba(OH)2 produces two moles of OH-.
Since the volume of the solution is 125 mL, we can convert it to liters:
Volume = 125 mL = 125/1000 L = 0.125 L
Now, we can calculate the number of moles of OH- using the hydroxide concentration and the volume:
Moles of OH- = [OH-] x Volume
Moles of OH- = 4.28 x 10^(-4) M x 0.125 L
Moles of OH- ≈ 5.35 x 10^(-5) moles
Since there is a 1:1 stoichiometric ratio between OH- and Ba(OH)2, the number of moles of Ba(OH)2 is the same as the moles of OH-.
Therefore, the number of moles of Ba(OH)2 dissolved in the solution is approximately 5.35 x 10^(-5) moles.
To find the mass of Ba(OH)2, we can use the molar mass of Ba(OH)2, which is calculated by adding the atomic masses of each element in the compound:
Molar mass of Ba(OH)2 = (atomic mass of Ba) + 2*(atomic mass of O) + 2*(atomic mass of H)
Using the atomic masses:
Molar mass of Ba(OH)2 = 137.33 g/mol + 2*(16.00 g/mol) + 2*(1.01 g/mol)
Molar mass of Ba(OH)2 = 171.33 g/mol
Now, we can calculate the mass of Ba(OH)2:
Mass of Ba(OH)2 = Moles of Ba(OH)2 x Molar mass of Ba(OH)2
Mass of Ba(OH)2 = 5.35 x 10^(-5) moles x 171.33 g/mol
Mass of Ba(OH)2 ≈ 0.00915 grams
Therefore, approximately 0.00915 grams of Ba(OH)2 have been dissolved in the 125 mL solution.