Ask Question
10 July, 07:39

How many grams of nh4cl (mm = 53.49 g/mol) must be added to 0.250 l of 0.375 m nh3to produce a buffer solution with ph = 9.45? (kb of nh3 is 1.8 * 10-5) ?

+1
Answers (1)
  1. 10 July, 10:58
    0
    When we have PH = 9.45 so, we can get POH from:

    PH = 14 - POH

    POH = 14 - 9.45

    = 4.55

    So by using H-H equation, we can get [NH4+]:

    POH = Pkb + ㏒ [B-/HB-]

    when Kb = 1.8 x 10^-5

    ∴Pkb = - ㏒ (1.8 x 10^-5)

    = 4.7

    by substitution:

    4.55 = 4.7 + ㏒[0.375]/[NH4]

    ∴[NH4] = 0.53 M

    now we need to get the moles of NH4 at 0.25 L

    moles NH4 = molarity * volume

    = 0.53 M * 0.25L = 0.1325 moles

    now, we can get the mass of NH4 when:

    Mass NH4 = moles NH4 * molar mass

    = 0.1325 moles * 53.49 g/mol

    = 7 g
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “How many grams of nh4cl (mm = 53.49 g/mol) must be added to 0.250 l of 0.375 m nh3to produce a buffer solution with ph = 9.45? (kb of nh3 ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers