What is the [OH-], [H30+], and pH of a solution containing 0.125 g NaOH in 0.500 L of solution?

Answer 1

ANSWER:
pH = 11.8
[OH-] = 6.25 x 10^-3
[H3O+] = 1.58 x 10^-12

EXPLANATION:
First convert the mass of NaOH to mols of NaOH. Then divide the mol NaOH by volume. This will show that 0.125g NaOH in .5L of solution is 6.25 x 10^-3 in molarity.

6.25 x 10^-3 M NaOH means that there is [OH-] = 6.25 x 10^-3 as well, because 1mol of NaOH releases one more of OH- when it ionizes.

From there use the following equations to solve for rest.
pH = -log[H+]
pH + pOH = 14
pOH = -log[OH-]

Answer 2

You've got grams of NaOH, so you can find moles of NaOH. This is useful, since you have L solution. Find your molarity by calculating moles NaOH over L solution.

Since NaOH is a strong base, it dissociates completely in aqueous solution. That means the concentration of your OH- ion is equal to the molarity of your NaOH solution. That's your [OH-].

To find your pH, first use the [OH-] to find pOH. Since pOH = -log [OH-], and pH + pOH = 14.00, you should be able to find your pH.

You know that [H30+] is the same as [H+], and pH = -log [H+], so just plug in your pH and take the antilog to find your [H+]