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10.0 mL of a 0.300M NH3 solution is titrated with a 0.100M HCl solution. Calculate the pH after the following additions of HCl.

  • Chemistry -

    I will do the third one.

    NH4OH: .010l*.3=.0030 moles
    HCl: .030*.1=.003 moles

    they are all react, no excess acid or base. So, pH=7

    the fourth one:
    moles base: .0030
    moles acid: .040*.1=.004, or an excess of acid of .001moles

    volume+ 10ml+40ml=.050 liters

    concen acid: .001/.050=.02M

    pH= -log(.02)= 1.70

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