[LucaZombini]

to a 1,16 dm3 NH3 solution (pH 11.68) are added 0,25614 dm3 of HCl 7,3% (density 1,185kg/dm3). calculate the pH. the result should be 8.65

here is my procedure

14-11.68 = 2.32
10^-2.32 = 4.78e-3
cNH3 = ((4,78e-3)^2)/(1,8*10^-5) = 1,26 mol/L
nNH3 = 1,16 * 1,26 = 1,47

nHCl = 0,256*1185*0,073/36.45 = 0,607

[OH-] = (1,8-10^-5)*(1,47-0,607)/0,607 = 2,55 * 10^-5
pOH = 4.59, so the pH i get is 9.4

[Borek]

If you expect us to help don't make as guess what the numbers are and what you are doing, at least explain "I am calculating this", "I am calculating that".

This is a buffer solution, not ammonia solution, right? So why do you calculate pH as if the solution contained only NH₃? NH₄⁺ didn't disappear, it is still in the solution and shifts the dissociation equilibrium.

Is there any equation that is commonly used for calculating buffers?

[chenbeier]

to a 1,16 dm3 NH3 solution (pH 11.68) are added 0,25614 dm3 of HCl 7,3% (density 1,185kg/dm3). calculate the pH. the result should be 8.65

here is my procedure

14-11.68 = 2.32
10^-2.32 = 4.78e-3
cNH3 = ((4,78e-3)^2)/(1,8*10^-5) = 1,26 mol/L
nNH3 = 1,16 * 1,26 = 1,47

nHCl = 0,256*1185*0,073/36.45 = 0,607

[OH-] = (1,8-10^-5)*(1,47-0,607)/0,607 = 2,55 * 10^-5
pOH = 4.59, so the pH i get is 9.4

1. Calculate the amount of ammonia. Use equation for weak base.
2.Calculate the concentration and amount of hydrochloric.
3. Do neutralization and calculate the amount of ammonium
4. Use Henderson Hasselbalch to calculate the new pH

[AWK]

The density of 7.3% HCl ~1.03. The stated density concerns ~38% HCl (at a temperature of 25°C). But this does not change the fact that there is another error in the data.
 For these data, the task is calculated correctly although a bit illegible.

[LucaZombini]

i'm re-writing it down here

o a 1,16 dm3 NH3 solution (pH 11.68) are added 0,25614 dm3 of HCl 7,3% IN WEIGHT (density 1,185kg/dm3). calculate the pH. the result should be 8.65

here is my procedure

calculating the [OH-] (which is equal to the NH4+ concentration)
14-11.68 = 2.32
10^-2.32 = 4.78e-3

using the Kb equation for NH3 to find the NH3 concentration
Kb = [NH4+][OH-]/[NH3]
1,8*10^-5 = (4,78E^-3)^2 /[NH3]

cNH3 = ((4,78e-3)^2)/(1,8*10^-5) = 1,26 mol/dm3
nNH3 = 1,16 * 1,26 = 1,47

MW HCl = 36,45 g/mol
cHCl = 1185*0,073/ 36.45 = 2,37 mol/dm3
nHCl = 2,37*0,25614 = 0,607

calculating the buffer solution
[OH-] = Kb* nNH3/nNH4+

[OH-] = (1,8-10^-5)*(1,47-0,607)/0,607 = 2,55 * 10^-5
pOH = 4.59, so the pH i get is 9.4

i got the fact that i forgot to add the NH4+ but the 2 values are too different

[Borek]

Now it is much more easy to follow. 9.4 is a correct answer for the data given.