Respuesta :
Answer:- 9.58
Solution:- We can calculate the pH for the given one using Handerson equation.
pH = pKa + log(base/acid)
Ammonia is the base and ammonium ion (ammonium chloride) is the acid.
We have 10.70 grams of ammonium chloride. Molar mass of ammonium chloride is 53.49 g/mol.
moles of ammonium chloride or ammonium ion = 10.70 g x (1mol/53.49 g)
= 0.2000 mol
We have 35.00 ml of 12 M ammonia solution. So, we can calculate the moles of ammonia also.
35.00 ml is 0.0350 L, So....
0.0350 L x (12 mol/1L) = 0.420 mol
Volume of the buffer is 1.00 liter, so the molarities for both the solutions would be same as their number of moles.
[base] = 0.420 M and [acid] = 0.200 M
We can calculate pKb from given kb value.
pKb = - log Kb
pKb = [tex]- log(1.80*10^-5)[/tex]
pKb = 4.74
Now we can calculate pKa as...
pKa = 14 - pKb
pKa = 14 - 4.74
pKa = 9.26
So, pH = [tex]9.26 + log(\frac{0.420}{0.200})[/tex]
pH = 9.26 + 0.322
pH = 9.58
The PH of the given buffer system that is prepared is gotten as; PH = 9.58
We are given;
Mass of NH4Cl = 10.7 g
Volume of NH3 = 35 mL = 0.035 L
Molarity NH3 = 12 M
Let us calculate the number of moles for both NH4Cl and NH3.
Number of moles of NH4Cl = mass/molar mass
Molar mass of NH3 = 53.49 g/mol
Thus;
Number of moles of NH4Cl = 10.7/53.49
Number of moles of NH4Cl; n_acid = 0.2 mol
Number of moles of NH3 = Volume × molarity
Number of moles of NH3 = 0.035 × 12
Number of moles of NH3; n_ base = 0.42 mol
Now, we are given the base dissociation constant of NH3 as;
kb = 1.8 × 10^(-5)
Formula for Pkb is;
Pkb = -log kb
Pkb = -log(1.8 × 10^(-5))
Pkb = 4.745
Pka = 14 - Pkb
Thus;
Pka = 14 - 4.745
Pka = 9.255
Formula for the PH here is;
PH = Pka + log(n_base/n_acid)
PH = 9.255 + log(0.42/0.2)
PH = 9.577
Approximately PH = 9.58
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