A buffer solution having a total volume of 0.50 L is prepared which has the foll
ID: 1025442 • Letter: A
Question
A buffer solution having a total volume of 0.50 L is prepared which has the following composition: [NH4Cl] = 0.25 M, [NH3] = 0.40 M. [Note: Kb of NH3 = 1.8 × 105 ]
(a) Identify the species that acts as the acid and base in this buffer.
(b) What role does the chlorine anion play?
(c) Calculate the pH of this buffer solution using the Henderson-Hasselbalch equation. Is there any underlying assumption being made when you use this equation? Please show your work.
(d) If you added 0.10 moles of HCl to the solution, what would the new pH be? Please show your work. (e) Based on your answer to question (a), determine the buffer range. Please show your work.
Explanation / Answer
(A) In this buffer ammonia acts as the base and ammonium chloride acts as the acid. NH4+ being the conjugate base
(b) chlorine gives the acidity to the buffer. NH4Cl dissociates in water to give ammonium and chlorine ions. the Cl-then associates with water to form HCl which dissociates to give H+ and Cl- ions.
(c)the henderson-hasselbalch equation for this is
pOH= pKb + log {[BH+]/[B]}
= -log(1.8 * 10-5) + log [0.25]/[0.40]
= 4.744 + (-0.204)
= 4.54
pH= 14- pOH = 14-4.54
pH= 9.46
(d) if you add HCl, it will dissociate into H+ and Cl-. the H+ will form H3O+ with water. concentration of HCl is 0.1/0.5(moles/vol)= 0.2M. so [H3O+]=0.2M
the ammonia will now react with these hydronium ions as follows
NH3 + H3O+ -> NH4+ + H2O
I 0.4 0.2 0.25
C 0.4-0.2 -0.2 +0.2
E 0.2 0 0.45
using the hendelsen-hasselbalch equation,
pOH= pKb + log {[BH+]/[B]}
= -log(1.8 * 10-5) + log [0.45]/[0.20]
=4.744 + 0.352
(e) to find the buffer range we need to find the pKa. we are given kb so pKb= -log Kb= 4.744
pKa+ pKb=14
pKa= 14-4.744= 9.25
the buffer range is usually +/- pKa
so, the buffer range anywhere is between 8.25 to 10.25
= 5.096
so pH= 14- pOH= 14-5.096
pH = 8.9
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