A particular acid [HA-] with pKa=5.64 is used to make a buffered solution with a pH of 5.28 ; the total salt concentration is 43 mM.  U?

A particular acid [HA-] with pKa=5.64 is used to make a buffered solution with a pH of 5.28 ; the total salt concentration is 43 mM. How many millimoles of conjugate base [A-] were used? Hint: Use the Henderson-Hasselbalch equation and this equation [HA] + [A-] = 43

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  • Bobby
    Lv 7
    1 month ago

    pH = pKa + log [A-] / [HA]

    5.28 = 5.64 + log [A-] / [HA] 

    log [A-] / [HA] = -0.36

    taking antilogs

    [A-] / [HA] = 10^- 0.36 =0.436515832    ,,, eq 1 

    [HA] + [A-] = 43 mM.....given 

    substituting in 1 

    [A-] / 43 - [A-] = 0.436515832

    [A-] = 0.436515832 (43 - [A-])

    [A-] = 18.77018079 - 0.436515832 [A-] 

    1.436515832 [A-] = 18.77018079

    [A-] = 13.07 mM

    [HA] = 29.93 mM

    testing 

    5.28 = 5.64+ log( 0.436515832 ) = 5.28 

    your answer [A-] = 13.07 mM

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