[raafiki]

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 800. mL of 1.255-M solution of NH3 in order to prepare a pH = 9.10 buffer?

I got 74.08 but it said that was wrong...



Also could you help with this one too?

What volume (to the nearest 0.1 mL) of 5.40-M HCl must be added to 0.700 L of 0.200-M K2HPO4 to prepare a pH = 7.80 buffer?

I got 8.7 but that was wrong:/

Um i was so lost on the first one that i dont remember what i did... I got it wrong three times and gave up but i used pH=pKa+log (base/acid)




well for the second one, I did  7.8=7.2+log(.14-5.4x/5.4x) using the same pH formula stated above. I got .14 by multiplyin .700L by .200 M












My table for it was:
HCl +  HPO4 <-------> Cl- +   H2PO4-
5.4x     .14                 0         0
-5.4x    -5.4x            +5.4x    +5.4x
   0        .14-5.4x         5.4x      5.4x

Thanks for any help

[Borek]

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 800. mL of 1.255-M solution of NH3 in order to prepare a pH = 9.10 buffer?

I got 74.08 but it said that was wrong...

Show how you got it.

[raafiki]

How many grams (to the nearest 0.01 g) of NH4Cl (Mm = 53.49 g/mol) must be added to 800. mL of 1.255-M solution of NH3 in order to prepare a pH = 9.10 buffer?

I got 74.08 but it said that was wrong...

Show how you got it.

Um i was so lost on the first one that i dont remember what i did... I got it wrong three times and gave up but i used pH=pKa+log (base/acid)

well for the second one, I did  7.8=7.2+log(.14-5.4x/5.4x) using the same pH formula stated above. I got .14 by multiplyin .700L by .200 M



My table for it was:
HCl +  HPO4 <-------> Cl- +   H2PO4-
5.4x     .14                 0         0
-5.4x    -5.4x            +5.4x    +5.4x
   0        .14-5.4x         5.4x      5.4x

[Borek]

Your approach to the second one is correct, just check your math, as 8.7 is not a solution to the (correct) equation that you wrote.