Question

In: Chemistry

Which of the following aqueous solutions are good buffer systems? . a. 0.15 M sodium hydroxide...

Which of the following aqueous solutions are good buffer systems?
.

a. 0.15 M sodium hydroxide + 0.29 M sodium bromide .

b. 0.31 M hydrofluoric acid + 0.20 M potassium fluoride .

c. 0.27 M nitric acid + 0.24 M potassium nitrate .

d. 0.16 M hydrocyanic acid + 0.18 M potassium cyanide .

e. 0.31 M ammonium nitrate + 0.30 M ammonia .

b.How many moles of sodium hydroxide would have to be added to 250 mL of a 0.406 M acetic acid solution, in order to prepare a buffer with a pH of 4.600?

moles

Solutions

Expert Solution

Acid buffer solution : It is a mixture of weak acid and its salt with strong base

HF + KF

CH3COOH + CH3COONa

HCN + NaCN

Basic buffer solution : It is a mixture of weak base and it's salt with strong acid

NH3 + NH4Cl

NH3 + NH4NO3

b. 0.31 M hydrofluoric acid + 0.20 M potassium fluoride .

d. 0.16 M hydrocyanic acid + 0.18 M potassium cyanide .

e. 0.31 M ammonium nitrate + 0.30 M ammonia .

b.

no of moles of CH3COOH = molarity * volume in L

                                           = 0.406*0.25 = 0.1015 moles

-----    CH3COOH(aq) + NaOH(aq) --------------> CH3COONa(aq) + H2O(l)

I --------    0.1015 ---------   x ----------------------        0

C-------   -x   -------------    -x -----------------------   +x

E ------ 0.1015-x ------- 0 ------------------------------ +x

         Pka = 4.74

         PH   = 4.6

         PH   = Pka + log[CH3COONa]/[CH3COOH]

        4.6 = 4.74 + logx/(0.1015-x)

logx/(0.1015-x)   = 4.6-4.74

logx/(0.1015-x)   = -0.14

x/(0.1015-x)    = 0.7244

x                    = 0.7244*(0.1015-x)

x   = 0.0426

The no of moles of NaOH = x   = 0.0426 moles >>>>answer


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