Question

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?

Answer 1

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 10^x 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.

Answer 2

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

Answer 3

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

Answer 4

i don't get the OH^-

Answer 5

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.

Answer 6

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

Answer 7

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.

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

pH = -log(H^+)
9.35 = -log(H^+).
solve for (H^+). <b>Do you know how to do that on the calculator? Punch in 9.35, change the sign to negative, hit the 10^x key. The (H^+) pops up</b>
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.

Answer 12

is it-417649 or -.667

Answer 13

I don't even know the problem anymore.
<b>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 10^x 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. </b>