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Ph=4.390 write the henderson-hasselbalch equation for a propanoic acid solution (ch3ch2co2h, pka=4.874). 4.390

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Answer:

a) [A⁻]/[HA] = 0.227b) [A⁻]/[HA] = 0.991c) [A⁻]/[HA] = 2.667Explanation:In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:CH₃CH₂CO₂H = HACH₃CH₂CO₂⁻ = 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] = [A⁻]/[HA]0.227 = [A⁻]/[HA](b) 4.87 = 4.874 + Log [A⁻]/[HA]-0.004 = Log [A⁻]/[HA] = [A⁻]/[HA]0.991 = [A⁻]/[HA](c) 5.30 = 4.874 + Log [A⁻]/[HA]0.426 = Log [A⁻]/[HA] = [A⁻]/[HA]2.667 = [A⁻]/[HA]

Explanation:

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