Question

A) Consider how best to prepare one liter of a buffer solution with pH = 4.94 using one of the weak acid/conjugate base systems shown here.

Weak Acid	Conjugate Base	Ka	pKa
HC2O4-	C2O42-	6.4 x 10-5	4.19
H2PO4-	HPO42-	6.2 x 10-8	7.21
HCO3-	CO32-	4.8 x 10-11	10.32

How many grams of the potassium salt of the weak acid must be combined with how many grams of the potassium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams potassium salt of weak acid =

grams potassium salt of conjugate base =

B) Design a buffer that has a pH of 3.75 using one of the weak acid/conjugate base systems shown below.

Weak Acid	Conjugate Base	Ka	pKa
HC2O4-	C2O42-	6.4×10-5	4.19
H2PO4-	HPO42-	6.2×10-8	7.21
HCO3-	CO32-	4.8×10-11	10.32

How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams sodium salt of weak acid =  g

grams sodium salt of conjugate base =  g

C) Consider how to prepare a buffer solution with pH = 9.52 (using one of the weak acid/conjugate base systems shown here) by combining 1.00 L of a 0.301-M solution of weak acid with 0.365 M potassium hydroxide.

Weak Acid	Conjugate Base	Ka	pKa
HNO2	NO2-	4.5 x 10-4	3.35
HClO	ClO-	3.5 x 10-8	7.46
HCN	CN-	4.0 x 10-10	9.40

How many L of the potassium hydroxide solution would have to be added to the acid solution of your choice?

Design a buffer that has a pH of 7.29 using one of the weak base/conjugate acid systems shown below.

Weak Base	Kb	Conjugate Acid	Ka	pKa
CH3NH2	4.2×10-4	CH3NH3+	2.4×10-11	10.62
C6H15O3N	5.9×10-7	C6H15O3NH+	1.7×10-8	7.77
C5H5N	1.5×10-9	C5H5NH+	6.7×10-6	5.17

D) How many grams of the chloride salt of the conjugate acid must be combined with how many grams of the weak base, to produce 1.00 L of a buffer that is 1.00 M in the weak base?

grams chloride salt of conjugate acid =

grams weak base =

Answer 1

First 2 questions has been solved here. kindly post the other questions separately.

Selecting the acid/conjugate base pair depends on the value of pKa 1 . This range should have the desired pH. So,

In a) The following pair would be used, pKa = 4.19 and the range of pH for which this buffer would be good = 3.19 to 5.19.

HC2O4-

C2O42-

Using Henderson - Hasselbalch equation:

pH = pKa + log { [conjugate base]/[acid] }

Here potassium salt of the weak acid and potassium salt of its conjugate base is taken,

log { [K₂C₂O₄]/[KHC₂O₄] } = 4.94 - 4.19 = 0.75

Or, log [K₂C₂O₄] - log [KHC₂O₄]    = 0.75

Given that [K₂C₂O₄] is = 1.0 M

So, log [K₂C₂O₄] = 0

And, log [KHC₂O₄] = 10 ^-0.75 = 0.178 M

Since, volume is 1 liter, moles = molarity (numerically)

moles of KHC₂O₄ = 0.178

mass of KHC₂O₄ = 0.178 x (128 g/mol)      = 22.76 g           .....Note that molar mass of KHC₂O₄ is = 128 g/mol

mass of K₂C₂O₄   = 1 x 166 g/mol    = 166 g     .....Note that molar mass of KHC₂O₄ is = 166 g/mol

Now, Note that effectively we are doing the following steps:

Step 1) choose the correct weak acid/conjugate base pair by checking its pKa and given pH range

Step 2) applying Henderson equation to get the concentration of acid.

         if concentration of conjugate base = 1.0 M, log [conjugate base ] =0 , as solved for a) above,

And volume = 1 litre , so molarity = moles (numerically)

that is; [salt of weak acid] = 10 ^{-(pH -pKa)} = moles of salt of weak acid

Step 3) amount in grams of salt of weak acid = molar mass of salt x 10 ^{-(pH -pKa)}

And      amount in grams of salt of conjugate base = 1 x molar mass of salt (if molarity given is 1.0 M)

So, for part b) Again , the pair is

HC2O4-

C2O42-

pH = 3.75 , directly, amount in grams of NaHC₂O₄= molar mass of salt x 10 ^{-(3.75 - 4.19)}

                                                                                    = (112 g/mol ) x 10^0.44   = 308.47 g
And,

                                     amount in grams of Na₂C₂O₄ = 1 x molar mass of salt = 134 g