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Question: $K_b$ of a weak base BOH is $10^{-5}$. 100 mL of 0.2 M BOH is mixed with 100 mL of 0.1 M HCl. What w...

KbK_b of a weak base BOH is 10−510^{-5}. 100 mL of 0.2 M BOH is mixed with 100 mL of 0.1 M HCl. What would be the pH of solution obtained by mixing BOH and HCl?

Answer

pH = 9

Explanation

Solution

Solution:

  1. Moles Calculation:

    • Moles of BOH = 0.1 L × 0.2 M = 0.02 moles
    • Moles of HCl = 0.1 L × 0.1 M = 0.01 moles

  2. Neutralization Reaction:

    BOH + HCl → BCl + H₂O

    Remaining BOH = 0.02 – 0.01 = 0.01 moles

    Moles of salt (BCl) formed = 0.01 moles

  3. Concentration in Final Volume:

    Total volume = 0.1 L + 0.1 L = 0.2 L

    [BOH] = 0.01 / 0.2 = 0.05 M

    [BCl] = 0.01 / 0.2 = 0.05 M

  4. Buffer pOH Calculation:

    Using Henderson-Hasselbalch for a weak base:

    pOH = pK_b + log([BCl] / [BOH])

    Given pK_b = –log(10−510^{-5}) = 5 and ratio = 1

    pOH = 5 + log 1 = 5

  5. pH Determination:

    pH = 14 – pOH = 14 – 5 = 9