Question

10. What is the [CH3COOH ]/[CH3COO– ] ratio in an acetic acid/sodium acetate buffer at pH = 4.90? (Ka = 1.8 x 10–5 )

a. 0.31

b. 0.70

c. 1.45

d. 2.41

e. 4.90

11. How many grams of NaCH3COO should be added to 250 mL of 0,100 M CH3COOH when the above solution is prepared?

a. 0.431 g

b. 2.95 g

c. 5.66 g

d. 1.43 g

e.11.9 g

Answer 1

10. Given

pH = 4.90 and Ka = 1.8 *10 ^-5

consider dissociation of acetic acid at equilibrium,

CH₃COOH ⇌ CH₃COO ^- + H⁺

now equilibrium constant Ka ={ [CH₃COO^– ] [ H+] } / [CH₃COOH] ........equation 1

from pH we can calculate concentration of H⁺

pH = - log₁₀ [ H⁺]

therfore [ H⁺] = antilog₁₀ [ -pH ]

= antilog₁₀ [ -4.90]

[ H⁺] = 1.25 * 10 ^-5

Now substituting value of Ka and H+ in eqution 1 we get

  1.8 *10 ^{-5 =} [CH₃COO^– ] * 1.25 * 10 ^-5 /   [CH₃COOH]

but in problem we have been asked ratio of [CH₃COOH] / [CH₃COO^– ] hence we rearrange above equation as follows:

[CH₃COOH] / [CH₃COO^– ] = 1.25 * 10 ^-5 /   1.8 *10 ^-5

[CH₃COOH] / [CH₃COO^– ] = 0.6944 which is nearly equal to 0.70 so option is b.

11.

Given

V = 250 ml = 0.250 l

C = 0.100 M

pH = 4.90

Ka = 1.8 *10-5

pKa of acetic acid = 4.75 from literature

Weight of CH3COONa ? in grams

We know Henderson–Hasselbalch equation as follows:

now substituting this values we get,

4.90 = 4.75 + log ₁₀ ([salt] /  [Acid] )

log ₁₀ [salt] /[CH₃COOH] = 4.90 - 4.75

[salt] /[CH₃COOH] = antilog 10 (0.15)

[salt]/ [CH₃COOH]   = 1.4125

[salt] = 1.413 * 0.100

[salt] = 0.1413 M

we know concentration = moles / volume

moles = conc * volume = 0.1413 * 0.250 = 0.035325 moles

mass of salt = number of moles * molecular weight of CH₃COONa

= 0.035325 * 82 = 2.8966 g nearly equal to 2.9 g so answer is b