The largest change in ph for an effective buffer occurs when the base is how many times as concentrated as the acid?
(a) 25
(b) 10
(c) 5
(d) 40
(e) 8

(B) 10

Explanation:

For the following reaction -

 HA  +  H₂O >  H₃O⁺   + A⁻

 The pH equation for the above reaction is written as -

pH  =  pKa + log ( [base] / [acid] )

Hence , for a buffer to be effective , the value of pH should be equal to the value of  pKa , i.e. ,

pH  =  pKa

Hence , the value of pH should be equal to the value of  pKa , the log vale should be equal to 1 , which is done , when ,  the base is ten many times as that of the concentrated as the acid .

So ,  correct option is B .

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.