Question

PPlease help me by checking part a, helping me do part b and c. Please show your work.

1. (4 points) A buffer solution is created with C2H5NH 4.7 x 104 and 0.012 M C2H5NH2. The Kb for C2H5NH2 1s -3 What is the concentration of C2H5NH3 in the buffer solution that has a pH of 11.50? PH-11.5 pOHpbtlg( CLHoNHa 4.7x10) a. poHこI4-11,6-2,50 2.60 3.3 0g 0. T 0183-1 6.o12) C2HりNHL Prb= =メ/0101 10 -る(X-0.0018 M) What is the concentration of the buffer solution if 0.0011 M HCl is added to the buffer in part a.? b. C2HONH2 0.0109nM (0.012-00011) C2HのJHst -し Why is the buffer a basic solution? Could this weak acid/base pair create an acidic buffer solution? c. basio biu neoaset onjuya teaiid avepresent uffor io formed Why or why not?

Answer 1

a)

Firstly, we will calculate pK_b.

pk = -log K = -log (4.7 x 10-4) = 3.33

Then we will find pOH= 14- pH = 14-11.50 = 2.50

Then, we will use Henderson Hasselbalch equation,

pOH = pky + log- Salt (C2H5NH7] 1 = pk + log Base C2H5NH2

Plugging in the values, we have,

[C2H5NH3] 2.50 = 3.33+ log 0.012M [C2H5NH3] log 0.012M = -0.83 C2H5NH3] _ 10-0.83 0.012M C2H5NH 1 = 0.012M X 10-0.83 = 1.775 x 10-3M 1.8 x 10-3M

Thus, your calculation is correct.

b)

HCl will convert the base to its salt.

C2H5NH2 + H+ + C2H5NH,

Thus, [C₂H₅NH₂] = 0.012 M - 0.0011 M = 0.0109 M

[C₂H₅NH₃⁺]=0.0018 M + 0.0011 M= 0.0029 M

c)

The buffer is a basic buffer because the pH is above 7.0.

Yes, this buffer can create an acidic buffer as well depending on concentration of species. For Acidic buffer pH<7 or pOH>7. This will give,

POH > 7 (C2H5NH3) →pky + log [C2H5NH2] [CH NH] 3.33+ log [C2H5NH2] C2H5NH3) > 3.67 log C,H5NH2] C2H5NH3 103.67 [C2H5NH2] C2H5NH3> 4.7 x 10 C2H5NH2]

If this condition is satisfied then the "buffer" will theoretically act as an acidic buffer.