What is the pH of 500 mL of water that had 0.00150 grams of NaOH added to it? The units are pH units with an error interval of 0.1 units.
0.00150 divided by 40NaOH 3.75e-5 divided that by 500= 7.5e-8 -log7.5e-8 =7.12 im lost on the last step
Not quite. Thanks for showing your work.
0.00150/40 = mols KOH and that's right. But that's divided by L (0.500) and not 500. That give you about 7.4E-5 and that is the (KOH) in the solution. Take the -log of that to give pOH and subtract from 14 to obtain the pH.
is the answer 9.86923
No but I would go with 9.87
I don't know what you're supposed to do with the error unit of 0.1.
To find the pH of a solution, you need to determine the negative logarithm of the concentration of hydrogen ions (H+) in the solution. In this case, you are trying to find the pH of water to which NaOH has been added.
Let's break down the steps:
Step 1: Calculate the molarity of NaOH in the water.
Given that you added 0.00150 grams of NaOH to 500 mL of water, you need to convert the mass of NaOH to moles and divide by the volume of water in liters (500 mL = 0.5 L).
0.00150 grams NaOH / 40 g/mol NaOH = 3.75e-5 moles NaOH
3.75e-5 moles NaOH / 0.5 L = 7.5e-5 M NaOH
Step 2: Calculate the concentration of OH- ions.
Since NaOH dissociates completely in water, the concentration of hydroxide ions (OH-) is the same as the molarity of NaOH.
[OH-] = 7.5e-5 M
Step 3: Calculate the concentration of H+ ions.
In water, the concentration of H+ ions is given by the concentration of hydroxide ions multiplied by the concentration of hydrogen ions in water, which is approximately 1e-7 M.
[H+] = (7.5e-5 M) * (1e-7 M) = 7.5e-12 M
Step 4: Calculate the negative logarithm of [H+] to find the pH.
Using the formula pH = -log[H+]:
pH = -log(7.5e-12) ≈ 11.12
Now, let's address the confusion you had with the last step. The final answer should be pH ≈ 11.12, not 7.12. The negative logarithm of 7.5e-12 is approximately 11.12, indicating that the pH of the solution is around 11.12.
Keep in mind that the given error interval for pH units in this case is 0.1 units. Hence, you can round the pH to the nearest 0.1 interval, resulting in a final pH of 11.1.