An analytical chemist is titrating 146.0mL of a 0.5400M solution of methylamine CH3NH2 with a 0.4300M solution of HIO3?

An analytical chemist is titrating 146.0mL of a 0.5400M solution of methylamine CH3NH2 with a 0.4300M solution of HIO3.Calculate the pH of the base solution after the chemist has added 51.1mL of the HIO3 solution to it.

The pKb of methylamine is 3.36.

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  • 6 months ago

    Initial moles CH3NH2 = 0.1460 L X 0.5400 mol/L = 0.07884 mol CH3NH2

    Moles HIO3 added = 0.0511 L X 0.4300 mol/L = 0.02197

    The HIO3 reacts with the base forming CH3NH3+. So, after the addition, moles CH3NH3+ = 0.02197 and moles CH3NH2 = 0.07884 - 0.02197 = 0.05687

    I'm going to use the Henderson-Hasselbalch equation to calculate the pH directly:

    pH = pKa + log [CH3NH2]/[CH3NH3+]

    pKa = 14.00 - pKb = 10.64

    pH = 10.64 + log (0.05687/0.02197) = 11.08

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