Consider the titration of 40.0 mL of 0.400 M HI by 0.300 M NaOH.
A) Calculate the pH after 34.7 mL of NaOH has been added.

B)At what volume (in mL) of NaOH added does the pH of the resulting solution equal 7.00? Include the units of mL in your answer.

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  1. millimoles HI = mL x M = 40.00 mL x 0.400 M = 16.00
    millimoles NaOH added = 34.7 mL x 0.300 M = 10.41
    ..........................HI + NaOH ==> NaI + H2O
    E........................ 5.59.........0..............10.41
    So what do you have? That's an excess of 5.59 millimoles HI which is a strong acid and 10.41 mmols of NaI (the salt of a strong base and a strong acid which is neutal in solution). The pH will be determined by the HI concentration. Convert that to pH. Post your work if you get stuck.

    B. For the pH to be 7.00 you will be at the equivalence point so you wantt the volume of the 0.300 M NaOH that will neutralize the 16.00 mmoles HI.
    M NaOH = millimoles HI/mL NaOH. YOu know M = 0.300 and you know mmols HI = 16.00. Solve for pH = M = mmoles/mL

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