Chemistry

When 23.5 mL of 0.12 M HCl are added to 50.0 mL of a 0.15 M solution of a weak monoprotic base, the pH of the solution is 10.00. What is the Kb of the weak base?

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  1. millimols HCl = mL x M = 23.5 x 0.12 = ?= approx 3
    millimols BOH = 50 x 0.15 = ? appox 7.5
    You need to use better accuracy..These are only approx.
    ............HCl + BOH ==> BCl + H2O
    I............3.........7.5...........0..........0
    C,,,,,,,,,-3.........-3.............+3.......+3
    E...........0.........4.2...........3

    You see you have a buffered solution consiting of a weak base and it's acid (BCl). Plug this into the Henderson-Hasselbalch equation, solve for pKa which is the only unknown. Then to get pKb use
    pKa + pKb = pKw = 14. You know pKw and pKa, solve for pKb.
    Post your work if you get stuck.

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  2. 23.5-ml(0.12 HCl) + 50-ml(0.15M BH:OH)
    => 0.0235(0.12) mole HCl + 0.050(0.15) mole BH:OH
    => 0.00282 mole HCl + 0.0075 mole BH:OH
    => (0.0075 – 0.00282) mole BH:OH + 0.00282 mole BH⁺:Clˉ
    => 0.00468 mole BH:OH + 0.00282 mole BH⁺:Clˉ
    => (0.00468 mole BH:OH/0.0735-L Soln) + (0.00282 mole BH⁺:Clˉ/0.0735-L Soln)
    => (0.0367M) BH:OH + (0.0384M BH⁺:Clˉ) => pH = 10 => pOH = 14 – 10 = 4 => [OHˉ] = 10ˉ⁴M

    ……… BH:OH ==> BH⁺ + OHˉ
    I ….. 0.0367M …….. 0.0384M ……….0
    C …….. -X ………………….. +X ……………+X
    E …..0.0367-X ……….0.0384+X……….10ˉ⁴M
    ……=0.0366M ………=0.0385M

    K(b) = [BH⁺][:OHˉ]/[BH:OH] = (0.0366)(0.0385)/(10ˉ⁴) = 1.05 x 10ˉ⁴

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    posted by Doc48

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