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Can someone please calculate the ph of the solutions below????????? ONLY 5 QUESTIONS?

1. 0.01 M HCl

2. 0.0010 M NaOH

3. 0.050 M Ca(OH)2

4. 0.030 M HBr

5. 0.150 M KOH

2 個解答

相關度
  • 4 年前

    You need to use:

    ph = - log M or pOH = -log M ( use this to basics)

    1. 0.01 M HCl

    pH = - log 0.01 / pH = 2

    2. 0.0010 M NaOH

    pOH = - log M / pOH = 3 / use this relation: pOH + pH = 14 / so, 3 + pH = 14 / pH = 11

    3. 0.050 M Ca(OH)2 ( this, you need to observe that rate of OH liberated, see: Ca(OH)2 ---> Ca^+2 + 2 OH- , so you have 2 OH- being liberated, so the concentration of OH- is gonna be 2 times the concentration of Ca(OH)2.

    so, pOH = - log M / pOH = - log 0,1 / pOH = 1,1 / pOH + pH = 14 / 1,1 + pH = 14 / pH = 12,9

    4. 0.030 M HBr

    pH = - log 0,030 / pH = 1,52

    5. 0.150 M KOH

    pOH = -log M / pOH = 0,823 / pOH + pH = 14 / pH = 13, 17

    Observation: when you are dealing with basics, you need to find the pOH fist, than you convert to pH.

    Second: you need to observe the stoichiometry. I mean, you need to observe how many mols are being formed of the basics or acid.

    Take care.

    Paulo, from Brazil.

  • 匿名
    4 年前

    Why don't you do it?

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