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

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:

pH = pka + Log [A⁻]/[HA]

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

4.23 = 4.874 + Log [A⁻]/[HA]

-0.644 = Log [A⁻]/[HA]

= [A⁻]/[HA]

0.227 = [A⁻]/[HA]

4.87 = 4.874 + Log [A⁻]/[HA]

-0.004 = Log [A⁻]/[HA]

= [A⁻]/[HA]

0.991 = [A⁻]/[HA]

5.30 = 4.874 + Log [A⁻]/[HA]

0.426 = Log [A⁻]/[HA]

= [A⁻]/[HA]

2.667 = [A⁻]/[HA]

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