You have a 750mL of a buffer solution that is composed of NH3 and NH4Cl with a Kb value 1.8x 10^5 and a pH 9.43...... How many mL 6.00M HCl can be added before the buffer is exhausted?
this is a NH3-NH4+ buffer
pKb = -log(1.8*10^-5) = 4.75
pKa = 14-4.75= 9.25
then..
pH = pKa + log(NH3/NH4+)
9.43 = 9.45 + log(NH3/NH4+)
10^(9.43 - 9.45 ) = (NH3/NH4+)
0.95499 = (NH3/NH4+)
assume total NH3 + NH4+ mix is = 1 M then
mol of NH3 + NH4+ = 0.75 mol
0.95499 = (NH3/NH4+)
NH3 = 0.95499 *NH4+
0.95499 *NH4++ NH4+ = 0.75
NH4+ = 0.75/(0.95499 +1) = 0.3836 mol
mol of NH3 --> 0.95499 *0.3836 = 0.36633
then
max amount of HCl possible --> mol of NH3
0.36633 mol of HCl
V = mol/M = (0.36633)/(6) = 0.061 = 61 mL max...
this assume total molarity of buffer = 1 M
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