Calculate [OH-] and pH for the following strong base solution: 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL
[Ca(OH)2] = 0.0105M; therefore, 10 mL diluted to 500 must have a concn of
0.0105M x (10/500) = ??M
Since there are two OH ions per molecule of Ca(OH)2, (OH^-) must be twice that, then pOH = -log(OH^-).
pH + pOH = pKw = 14; use that to calculate H+.
Yes, that's correct but I wouldn't round it to 11 but keep it as 10.62. (10.62 has two significant figures --besides the 10 which comes from the log part--and you are allowed two--and I would keep 10.62.
Oh, strong base solutions... they always bring back memories of high school chemistry. Alright, let's tackle this question with a smile!
To calculate [OH-], we first need to find the number of moles of Ca(OH)2 present in the 10.0 mL solution.
Moles of Ca(OH)2 = (0.0105 mol/L) * (0.010 L) = 0.000105 mol
Next, we'll calculate the concentration of [OH-]. Since Ca(OH)2 dissociates into two OH- ions per molecule, we'll have twice the number of moles for OH-.
[OH-] = (0.000105 mol) / (0.5 L) = 0.00021 M
Now, let's move on to pH. Since we're dealing with a strong base, we can jump directly to pOH and then convert it to pH.
pOH = -log[OH-] = -log(0.00021)
pOH ≈ 3.68
Finally, we can calculate pH using the equation pH + pOH = 14.
pH = 14 - pOH = 14 - 3.68
pH ≈ 10.32
So, in this strong base solution, [OH-] is approximately 0.00021 M, and the pH is around 10.32. Remember, laughter is the best pH-balancer!
To calculate the [OH-] and pH for a strong base solution, we can use the concentration of the base and the volume of the solution.
Step 1: Calculate the number of moles of Ca(OH)2
First, we need to find the number of moles of Ca(OH)2 in the solution. We can use the formula: moles = concentration × volume
moles = 0.0105 M × 10.0 mL = 0.105 mmol
Step 2: Calculate the concentration of hydroxide ions [OH-]
Since Ca(OH)2 completely dissociates in water, there are two moles of OH- ions for every mole of Ca(OH)2. Hence, we can double the number of moles of Ca(OH)2 to find the number of moles of OH- ions.
moles of OH- = 2 × 0.105 mmol = 0.21 mmol
Next, we need to calculate the concentration of OH- ions in the final solution. We divide the moles of OH- ions by the final volume of the solution.
[OH-] = (moles of OH-) / (final volume)
= 0.21 mmol / 500.0 mL
= 0.00042 M
Step 3: Calculate the pH of the solution
pOH = -log[OH-]
= -log(0.00042)
= 3.38
Since pH + pOH = 14, we can find the pH:
pH = 14 - pOH
= 14 - 3.38
= 10.62
Therefore, the [OH-] is 0.00042 M and the pH is 10.62 for the given strong base solution.
Thank you so much! I would really appreciate it!
[OH-]= 4.2x10^-4 M
pH=10.6 or 11
Are these answers correct?