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Posted by on Wednesday, May 7, 2008 at 9:35pm.

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 - , Wednesday, May 7, 2008 at 9:41pm

    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 - , Wednesday, May 7, 2008 at 10:52pm

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

  • chemistry - , Wednesday, May 7, 2008 at 11:01pm

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

  • chemistry - , Wednesday, May 7, 2008 at 11:04pm

    i don't get the OH^-

  • chemistry - , Wednesday, May 7, 2008 at 11:19pm

    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 - , Wednesday, May 7, 2008 at 11:21pm

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

  • chemistry - , Wednesday, May 7, 2008 at 11:28pm

    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 - , Wednesday, May 7, 2008 at 11:32pm

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

  • chemistry - , Wednesday, May 7, 2008 at 11:37pm

    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 - , Wednesday, May 7, 2008 at 11:39pm

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

  • chemistry - , Wednesday, May 7, 2008 at 11:54pm

    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 - , Wednesday, May 7, 2008 at 11:59pm

    is it-417649 or -.667

  • chemistry - , Thursday, May 8, 2008 at 12:09am

    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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