Assume 25°C. Use pKa values from refference text. 0.01 moles of ammonia (NH_3) a
ID: 1023983 • Letter: A
Question
Assume 25°C. Use pKa values from refference text.
0.01 moles of ammonia (NH_3) are added to 1 liter of pure water. Determine the final pH and [NH_4^+] and [NH_3] in the solution? How does pH and [NH_4^+] and [NH_3] change w-hen 0.004 moles of HCl, a strong acid, are added to the solution described in part (a)? HCl is added from a concentrated stock solution, so the volume change is negligible. Determine the concentration of [NH_3] and [NH_4^+] in a pH 7.0 solution in equilibrium with an atmosphere with partial pressure of NH_3(g) of 106-7 atm. The Henry's constant (K_H) for NH_3 is 10^1.8 mol Ls6-1 atm^-1.Explanation / Answer
a)
Use following steps
Find the concentration of NH3 in the solution
[NH3]= 0.01 /1.0 = 0.01 M
Calculate ka value of NH3 and then use ICE to get [OH-] , [NH4+] and [NH3] at equilibrium
Pka = 14- pkb
= 14-9.3=4.7
Ka = Antilog 4.7= 1.992 E-5
ICE
NH3 + H2O --- > NH4+ + OH-
I 0.01 0 0
C -x +x +x
E (0.01 -x) x x
Ka = 1.992 E-2 = x2/ 0.01-x
Since x is negligible,
1.992 E-5 = x2 / 0.01
x = 0.000446684
[OH-] = [NH+]= 0.000446684
pOH = - log [OH-]= - log 0.000446684
=3.35
pH = 14-pOH = 10.65
[NH3]= 0.01 - 0.000446684 = 0.009553316 M
b)
Initial concentration of NH3= 0.009553316 M
[NH4+]= 0.000446684 M
One mole of HCl converts one mole of NH3 to NH4+ therefore,
Volume is one L so the concentration = moles
Final mole/concentration of NH3 = 0.009553316 – 0.004 = 0.00555 M
Final concentration of NH4+ = 0.000446684 + 0.004 =0.00445 M
c)
Find out the concentration of NH3 using Henrys constant equation
C = kH x p
Here kH is the henrys law constant, p is the pressure
C = 101.8 x 1.0E-7
[NH3]=6.31E-6
Use Henderson Hasselbalch equation
pH = pka + log [NH3]/[NH4+]
7.0 = 9.3 + log ([6.31 E-6]/[NH4+]
[NH4+]= 0.001259 M
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