You are asked to bring the ph of 0.500l of 0.550 m nh4cl to 7.00. which of the following solutions would you use: 12.0 m hcl or 12.0 m nh3? how many drops (1 drop 0.05 ml ) of this solution would you use?

First calculate the pH of the NH4Cl solution. 

NH4+ + H2O ==> H3O+ + NH3 

The Ka for this reaction is Kw / Kb for NH3 (which is 1.8 x 10^-5). 

Ka NH4+ = (1 x 10^-14) / (1.8 x 10^-5) =5.6 x 10^-10 

Construct an "ICE" chart: 

. .[M] .NH4+ + H2O => H3O+ +NH3 
.initial. 0.550 . . . . . . . . . .0 . . . . .0 
.change ..-x . . . . . . . . . . . x . . . . .x . 
..final . .0.550-x . . . . . . . . x . . . . .x 

Ka = (x)(x) / (0.550-x) 

Since the x is small compared to 0.550, we ignore it in the denominator and have 

Ka = x^2 / 0.550 = 5.6 x 10^-10 
x = 1.8 x 10^-5 = [H3O+] 
pH = -log [H3O+] = -log (1.8 x 10^-5) = 
4.74. 

This solution is acidic. To get to a pH of 7.00, we will have to add base (12.0 M NH3). 

Adding NH3 to NH4+ will produce a buffer system with NH4+ being the acid part of the buffer and NH3 being the basic component. The pH of buffers is best calculated using the Henderson-Hasselbalch equation: 

pH = pKa + log ([base]/[acid]) or 
pH = pKa + log ((moles base) / (moles acid)) 

7.00 = pKa NH4+ + log ((moles base) / (moles acid)) 
7.00 = -log (5.6 x 10^-10) + log ((moles NH3) / (moles NH4+)) 
7.00 = 9.25 + log ((moles NH3) / (moles NH4+)) 
-2.25 = log ((moles NH3) / (moles NH4+)) 
10^-2.25 = ((moles NH3) / (moles NH4+)) = 0.00562 

The initial moles of NH4+ is 
M NH4+ x L NH4+ = (0.550)(0.500) = 
0.275 moles NH4+ 

So now we need to add NH3 until the ratio of moles NH3 / moles NH4 = 0.00562. 

Since adding NH3 won't cahnge the moles of NH4+, then 

0.00562 = moles NH3 / 0.275 
moles NH3 = (0.00562)(0.275) = 0.00155 moles NH3 

moles NH3 = M NH3 x L NH3 
0.00155 = (12.0)(L NH3) 
L NH3 = 0.00155 / 12.0 = 0.000129 L NH3 = 0.129 mL NH3 

Yes she is correct. the answer is completely correct

Cause its being chemically applied?