[cloud5]

Please help me to solve this... 

1. A buffer solution was prepared from a mixture of 150 cm³ of 0.80 mol dm^-3 NH₃ and 50cm³ of 1.60 mol dm^-3 NH₄NO₃.

a. Calculate the pH of this buffer solution.
b. What will be the pH of the resulting solution if 20.0 cm³ of 0.05 mol dm^-3 H₂SO₄ is added to the buffer solution?

[AWK]

http://www.chemicalforums.com/index.php?topic=33740.msg129388#msg129388
Read rule, especially rule no. 4

[cloud5]

I don't know I'm doing it correctly... I did it this way: Given K_b = 1.8x10^-5

a. pOH=pK_b-log ([base])/([salt])
          = -log(1.8x10^-5)-log (0.80)/(1.60)
          = 5.04
   
    pH= 14-5.04
        = 8.96

b. Number of moles of H⁺ added
    =(0.020)(0.05)(2)
    =0.002

    Number of moles of NH₃
    =(0.150)(0.80)-0.002
    =0.118

    Number of moles of NH⁺₄
    =(1.60)(0.050)+0.002
    =0.082

   pOH=4.74-log (0.118)/(0.082)
         =4.58

   pH=14-4.58
       =9.42

[Jorriss]

Are you familiar with the Henderson-hasselbalch equation?

pH=pKa + log (A-/HA)

[Borek]

Are you familiar with the Henderson-hasselbalch equation?

pH=pKa + log (A-/HA)

Obviously s/he is:

pOH=pK_b-log ([base])/([salt])

At least to some extent

[Jorriss]

Are you familiar with the Henderson-hasselbalch equation?

pH=pKa + log (A-/HA)

Obviously s/he is:

pOH=pK_b-log ([base])/([salt])

At least to some extent

lol, I'll be honest, I barely read the post as I'm at work. I just threw that out there when someone said buffer

[cloud5]

Please check my answer... I don't know this is correct or not... 

I don't know I'm doing it correctly... I did it this way: Given K_b = 1.8x10^-5

a. pOH=pK_b-log ([base])/([salt])
          = -log(1.8x10^-5)-log (0.80)/(1.60)
          = 5.04
   
    pH= 14-5.04
        = 8.96

b. Number of moles of H⁺ added
    =(0.020)(0.05)(2)
    =0.002

    Number of moles of NH₃
    =(0.150)(0.80)-0.002
    =0.118

    Number of moles of NH⁺₄
    =(1.60)(0.050)+0.002
    =0.082

   pOH=4.74-log (0.118)/(0.082)
         =4.58

   pH=14-4.58
       =9.42

[Jorriss]

You have to take into account a change in concentration when you mix the two solutions together. I didn't see if you did that.

[cloud5]

Can you show me the correct formula? I'm not good in chemistry, sorry...

You have to take into account a change in concentration when you mix the two solutions together. I didn't see if you did that.

[Jorriss]

I'll make up some new numbers.

But suppose you have 100 mL of 1 M Ammonia and 100 mL of 1 M Sodium Chloride.

If you mix these together your new solution has a volume of 200 mL, the combination of the individuals. But even though your volume has changed, the number of moles of ammonia and NaCl added, have not.

So if you had 100 mL of 1 M ammonia, you had .1 moles of ammonia.

The equation for dilutions is:

M1V1=M2V2 (this just sets moles equal to eachother).


(1M)(100mL)=M2(200mL)

M2= .5M


Apply the same principle to your problem. The new concentrations likely won't be, I believe they were, .8 and 1.60 molar respectively, but some new concentration based on your new volume. By chance, the problem may work out the same, but it should be noted.

[Borek]

By chance, the problem may work out the same, but it should be noted.

It is not by chance.

C_A^- = n_A^- / V

C_HA = n_HA / V

C_A^- / C_HA = (n_A^- / V) / (n_HA / V) = n_A^- / n_HA

Volume cancels out.

But in general you are right that omitting effect of dilution is a bad idea, unless you know what you are doing.

[Jorriss]

By chance, the problem may work out the same, but it should be noted.

It is not by chance.

C_A^- = n_A^- / V

C_HA = n_HA / V

C_A^- / C_HA = (n_A^- / V) / (n_HA / V) = n_A^- / n_HA

Volume cancels out.

But in general you are right that omitting effect of dilution is a bad idea, unless you know what you are doing.

Well, yeah, by chance was poor wording but I didn't want to say its ok to skip it because I don't believe he did it on purpose.

[AWK]

a.
pH= 14-5.04
        = 8.96

b.
   pH=14-4.58
       =9.42

How ph can increase after addition of strong acid?

[cloud5]

Oh yeah! It makes sense! Thanks for pointing that out. 

How ph can increase after addition of strong acid?