Question

Write the Henderson-Hasselbalch equation for a propanoic acid solution ( CH3CH2CO2H , pKa=4.874 ) using the symbols HA and A− , and the given pKa value for propanoic acid in the expression. pH= 4.874 + Log [A-]/[HA] Using the equation to calculate the quotient [A−] / [HA] at three different pH values.(a) pH = 4.23(b) pH = 4.87(c) pH = 5.3

Answer 1

Answer:

a) [A⁻]/[HA] = 0.227

b) [A⁻]/[HA] = 0.991

c) [A⁻]/[HA] = 2.667

Explanation:

In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:

• CH₃CH₂CO₂H = HA

• CH₃CH₂CO₂⁻ = A⁻

pH = pka + Log [A⁻]/[HA]

pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]

• (a)

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

$10^{-0.644}$ = [A⁻]/[HA]

0.227 = [A⁻]/[HA]

• (b)

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

$10^{-0.004}$ = [A⁻]/[HA]

0.991 = [A⁻]/[HA]

• (c)

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

$10^{0.426}$ = [A⁻]/[HA]

2.667 = [A⁻]/[HA]