Chemistry

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Calculate the pH of the 0.20 M NH3/0.25 M NH4Cl buffer.
I calculated the pH to be 9.15

What is the pH of the buffer after the addition of 10.0 mL of 0.10 M HCl to 72.5 mL of the buffer?

  • Chemistry -

    10 mL x 0.1M HCl = 1 mmols.
    72.5 mL x 0.2M = 14.5 mmols NH3.
    72.5 mL x 0.25M = 18.125 mmols NH4^+. You can round to th correct number of significant figures when you finish the calculation.

    .........NH3 + H^+ ==> NH4^+
    I........14.5...0.......18.125
    add.............1.............
    C.......-1.....-1........+1
    E.......13.5....0........19.125

    pH = pKa + log(13.5/19.125) = ?

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