Question

How many mL of a 4.00 M HCl solution must be added to 250 mL of a 0.250 M NH₃ solution to make a buffer with pH = 9.10? Look up K_a or K_b in a suitable source.

Answer 1

no of moles of NH3   = molarity *volume in L

                                  = 0.25*0.25   = 0.0625 moles

no of moles of HCl   = molarity *volume in L

                                   = 4*v = 4v moles

POH = 14-PH

         = 14-9.10

         = 4.9

------------- NH3(aq) + HCl(aq) ------------------> NH4Cl(aq)

I ------------- 0.0625 ---- 4v --------------------------- 0

C-------------- -4v ------- -4v --------------------------- 4v

E------------ 0.0625-4v --- 0 -------------------------- 4v

Pkb = 4.75

POH   = PKb + log[NH4Cl]/[NH3]

4.9   = 4.75 + log4v/(0.0625-4v)

log4v/(0.0625-4v) = 4.9-4.75

log4v/(0.0625-4v)     = 0.15

4v/(0.0625-4v)      = 1.4125

4v       = 1.4125(0.0625-4v)

v = 0.00915

the volume of HCl   = 9.15ml >>>>>answer