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Calculate the pH of the resulting solution if 30.0 mL of 0.300 M HCl(aq) is added to
(a) 35.0 mL of 0.300 M NaOH(aq).
(b) 40.0 mL of 0.350 M NaOH(aq).

  • chemistry - ,

    I will do one.

    first: moles HCL=.030*.3=.009 moles
    moles base=.035*.3=.015 moles

    so you have excess of .006 moles OH

    pOH= -log (.006/.65)
    then pH= 14-pOH

  • chemistry - ,

    Why did you divide by .65?

  • chemistry - ,

    Wait so, you are getting 0.65 from the total mL but shouldnt this be 65/1000 or 0.065?

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