help! deadline tomorrow )): 2. It is desired to prepare 200 ml of a buffer solution of pH 4.5 from acetic acid, CH3COONa and its salt, sodium actetate, CHCOONa. For CH3COOH, pKa = 4.74. Calculate:

Question

2. It is desired to prepare 200 ml of a buffer solution of pH 4.5 from acetic acid, CH3COONa and its salt, sodium actetate, CHCOONa. For CH3COOH, pKa = 4.74. Calculate:

a. the volumes of 0.20M CH3COOH and 0.25M CH3COONa

b. the mass of CH3COONa that must be added to 200ml of 0.20M CH3COOH

Answer 1

Use the henderson equation:

pH=pKa+log(A-/HA)

4.5=4.74+log(x)

Plug in the concentrations of each thing:

4.5=4.74+log(0.2/0.25)

Now, say that the volumes need to add to 200 mL (0.2 L), so that V_acid+V_base=0.2 L

Make one volume x, and the other 0.2-x, so that

0.2=(x)+(0.2-x)

4.5=4.74+log(0.2(0.2-x)/0.25x)

x=0.116326

So V of 0.2M base is 0.2-x=0.083674 L

V of 0.25M acid is x=0.116326 L

Check if the moles result in the pH

pH=pKa+log(mol base / mol acid)

mol base = 0.083674L (0.2 mol / L)

mol acid = 0.116326L (0.25 mol / L)

4.74+log (0.083674(0.2) / 0.116326(0.25))

=4.500003723

Yeah, it gives the pH of 4.5

So

0.083674 L = 83.674 mL of CH3COONa

and

0.116326 L = 116.326 mL of CH3COOH

Answer 2

Henderson Hasselbalch: pH = pK_A + log ([A^-]/[HA])

You know pH, pKa = logK_a

a) Let's call sodium acetate the base, B, [A^-] and acetic acid the acid, A [HA] (M stands for molarity)

You know that 200 = V_A + V_B from addition of volumes

You know that the concentration of A is V_AM_A0/200 and B is V_BM_B0/200

You can solve for M_B/M_A from HH relation and you know that it is also V_BM_B0/V_AM_A0 You now have two equations for two unknowns. M₀ means concentrations of solutions you are mixing (known)

b) Here you use the same M_B/M_A from HH relation. M_A is .2 Molar --> solve for M_B

Now it is a problem of the grams of sodium acetate you need to make 0.2 liters of M_B molar solution.

0,2liters * M_B * (Formula Mass of Sodium acetate in g/mole)