# chemistry

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a pH meter gives a readout of 9.35 when the probe is dipped into an aqueous solution containing the strong base NaOH. What is the molarity of this solution with the respect to sodium hydroxide?

• chemistry -

pH = -log(H^+)
9.35 = -log(H^+).
solve for )H^+). Do you know how to do that on the calculator. Punch in 9.35, change the sign to negative, hit the 10x key. The (H^+) pops up
Then (H^+)(OH^-) = Kw.
You know H and Kw. Solve for (OH^-)

There is an easier way, I think, to do it.
9.35 is pH. Then pOH = 14-9.35.
Then pOH = -log(OH^-).
Substitute for pOH, and solve for OH^-. Much simpler, I think.
Check my work.

• chemistry -

ok since 14-9.35=4.65
what do i do after this...

• chemistry -

pH = -log(OH^-)
4.65 = -log(OH^-)
Just follow the above instructions to convert this number to OH^-

• chemistry -

i don't get the OH^-

• chemistry -

The question gives you the pH and asks you to calculate the concentration of the hydroxide ion. (OH^-) is read as "the concentration of the hydroxide ion". Actually it is NaOH which the problem uses but since the NaOH is 100% ionized in solution, then (NaOH) and (OH^-) are the same. In step 1 you converted pH to pOH. Now you want to convert pOH to (OH^-) = (NaOH). And you do it the same way you converted pH = 9.35 to (H^+) in this same post. In fact, I gave specific instructions for how to do that.

• chemistry -

this is a hard one for me...what do i do with 4.65

• chemistry -

You convert it to (OH^-) by using the formula 4.65 = -log(OH^-). And you do that EXACTLY the same way as you converted 9.35 pH to (H^+) in this same post. Go back up to that point and read how you are to do that on your computer, then remember how to do it.

• chemistry -

oh..i didn't convert 9.35 into anything though..cause that's the number they gave me

• chemistry -

Yes, you did. At least I told you how. Go back and look to see how I told you to convert it to (H^+).

• chemistry -

all i did was since the pkw is 14..i just subtracted that from the 9.35 to give me 4.65

• chemistry -

pH = -log(H^+)
9.35 = -log(H^+).
solve for (H^+). Do you know how to do that on the calculator? Punch in 9.35, change the sign to negative, hit the 10x key. The (H^+) pops up
Then (H^+)(OH^-) = Kw.
You know H and Kw. Solve for (OH^-)

There is an easier way, I think, to do it.
9.35 is pH. Then pOH = 14-9.35.
Then pOH = -log(OH^-).
Substitute for pOH, and solve for OH^-. Much simpler, I think.
Check my work.

• chemistry -

is it-417649 or -.667

• chemistry -

I don't even know the problem anymore.
pOH = -log(OH^-)
4.65 = -log(OH^-)
-4.65 = log(OH^-)
So we want to take the antilog of both sides. The antilog of -log(OH^-) is just (OH^-). The antilog of -4.65 is done this way on your calculator.
Punch in 4.65, change the sign to negative (or punch in -4.65 at the beginning), then find the button that is marked 10x and hit it. The answer of 2.238 x 10-5 will pop up (which I would round to 2.24 x 1o-5. That is the answer. That is the sodium hydroxide, the NaOH, concentration.

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