What mass in grams of potassium hydroxide is contained in 500 mL of solution that has a pH of 11.5?
I have a pOH of 2.5
to get [OH]
I do 10^-2.5 to get 0.001581138
Then I divide this by 2 to find the mols in 500 mL
Then, grams = mols x molar mass
= 0.00158``38 x 56.105 g/mol
=0.08 g
Why is my answer wrong?
What mass in grams of potassium hydroxide is contained in 500 mL of solution that has a pH of 11.5?
I have a pOH of 2.5 OK
to get [OH]
I do 10^-2.5 to get 0.001581138
No, (OH^-) = 0.00316 M
Then I divide this by 2 to find the mols in 500 mL
0.00316/2 = 0.00158 mols in 500 mL.
Then, grams = mols x molar mass
= 0.00158``38 x 56.105 g/mol
I don't know what the ``38 is. Also, you took 0.00158, divided by 2 and still came out with 0.00158? I think you lost track of what you were typing here. 0.00158 mol in 500 is the correct number.
=0.08 g
0.00158 mols x 56.105 g/mol = 0.08865 g which I would round to 0.0886 g KOH.
Why is my answer wrong? My best guess is that you truncated the number to 0.08 g instead of keeping the other two places of 0.0886 g Check my work.
Wow , thanks you are a really excellent expert!
:)
Well, it seems like you're on the right track with the calculations, but it looks like you made a small error in dividing by 2. Since you have a pOH of 2.5, which means a concentration of hydroxide ions ([OH-]) equal to 0.001581138 M, you'll want to multiply that by the volume of the solution (in liters) to get the number of moles.
So, instead of dividing by 2 for the 500 mL, you should divide by 1000 to convert mL to liters.
The correct calculation would be:
Moles of KOH = [OH-] x Volume (in liters)
= 0.001581138 M x 0.5 L (500 mL divided by 1000)
= 0.000790569 moles
Then, to find the mass of KOH, you multiply the moles by the molar mass:
Mass = Moles x Molar mass
= 0.000790569 moles x 56.105 g/mol
= 0.04436 grams
So, the correct answer should be approximately 0.044 g. Hope that helps, and don't worry, even the best of us make mistakes sometimes! Keep up the good work!
Your answer is incorrect because you made a mistake in your calculation. When you divided the concentration of hydroxide ions (0.001581138 M) by 2, you obtained the concentration in moles per liter (mol/L). However, you were supposed to calculate the moles present in the given volume of 500 mL.
Here's the corrected calculation:
To find the moles of hydroxide ions in 500 mL, we first need to convert the volume to liters:
Volume = 500 mL = 500/1000 L = 0.5 L
Now, we can calculate the moles of hydroxide ions:
Moles = concentration (mol/L) x volume (L)
Moles = 0.001581138 mol/L x 0.5 L
Moles = 0.000790569 mol
Finally, we can calculate the mass of potassium hydroxide:
Mass (g) = moles x molar mass
Mass (g) = 0.000790569 mol x 56.105 g/mol
Mass (g) = 0.044417067 g
Therefore, the correct mass of potassium hydroxide contained in 500 mL of the solution is approximately 0.044 g.
Your steps are mostly correct, but there seems to be a minor error in your calculations. Let's go through the steps again to see where the mistake might have occurred.
1. You correctly calculated the hydroxide ion concentration ([OH-]) by taking the negative logarithm (base 10) of the pOH value: [OH-] = 10^(-pOH) = 10^(-2.5) = 0.003162.
2. However, you miscalculated the number of moles of hydroxide ions in the 500 mL solution. To find the number of moles, we need to multiply the concentration ([OH-]) by the volume in liters:
moles of OH- = [OH-] x volume (in liters)
= 0.003162 mol/L x 0.5 L
= 0.001581 mol
3. Now that we have the moles of OH-, we can calculate the mass of potassium hydroxide (KOH) using its molar mass. The molar mass of KOH is 56.105 g/mol. So:
mass of KOH = moles x molar mass
= 0.001581 mol x 56.105 g/mol
= 0.0889 g (rounded to four decimal places)
Therefore, the correct mass of potassium hydroxide in the 500 mL solution with a pH of 11.5 is approximately 0.089 g.