Question

Question 9 of 9 Submit How many moles of NaOH need to be added to 200.0 mL of a 0.200 M solution of HF to make a buffer with a pH of 3.30? (Ka for HF is 6.8 x 10-4) | mol 7 8 +/- 1 +/- 0 x 100

Answer 1

no of moles of HF   = molarity * volume in L

                              = 0.2*0.2 = 0.04moles

Ka   = 6.8*10^-4

Pka   = -logKa

         = -log(6.8*10^-4)

        = 3.17

PH = 3.3

HF(aq) + NaOH(aq) -----------> NaF(aq) + H2O(l)

I     0.04             x                               0

C      -x              -x                               +x

E    0.04-x          0                                +x

            PH    = Pka + log[NaF]/HF]

          3.3    = 3.17 + logx/(0.04-x)

logx/(0.04-x)   = 3.3-3.17

logx/(0.04-x)      = 0.13

x/(0.04-x)          = 1.35

x       = 1.35*(0.04-x)

x     = 0.023

The no of moles of NaOH = x = 0.023 moles