Chemistry

How many grams of dry NH4Cl need to be added to 2.10L of a 0.600M solution of ammonia,NH3 , to prepare a buffer solution that has a pH of 9.00? Kb for ammonia is 1.8x10^-5.

I just wanted to make sure I dd this write, can you please check my work?
Thank you!!!!

I found the pKa which was 9.255..
then I used that and plugged it in the Hasselbalch (i think that's what it's called) equation.
I got .92592....M for NH4Cl and i used that Molarity and multiplied it by the liters.
so the mass was 1.94 moles of NH4Cl
I then used that to find the grams and my answer came out to be 104.02 grams of NH4Cl

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  3. 👁 5,808
  1. Would you show your work and let me check it? It's the Henderson-Hasselbalch equation.

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  2. I keep getting about 1.08M for NH4Cl.

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  3. I got 9.25527 for the pKa
    then for the Henderson-Hasselbalch equation:
    9.00=9.25527+log(.600/acid)
    10^(-.25527)=(.600/acid)
    .5556/.600M=acid
    acid=.925925...M
    molarity (.925925)*2.10L=

    1.94 moles of NH4Cl
    1.94 moles *53.4 grams=104.02g of NH4Cl

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  4. I think you made an algebra error from line 4 to line 5 in your work.

    I got 9.25527 for the pKa
    then for the Henderson-Hasselbalch equation:
    9.00=9.25527+log(.600/acid)
    10^(-.25527)=(.600/acid)correct to this line. Then
    0.5556 = 0.6/acid and
    acid = 0.6/0.5556 = 1.08 which is the reciprocal of your number)


    .5556/.600M=acid
    acid=.925925...M
    molarity (.925925)*2.10L=

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  5. yes you are totally right.
    Thank you so much!!!!!!!

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