Write the Henderson-Hasselbalch equation for a solution of propanoic acid (CH3CH2CO2H, pKa = 4.874) using HA, A– and the pKa value given in the expression.
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Write the Henderson-Hasselbalch equation for a solution of propanoic acid (CH3CH2CO2H, pKa = 4.874) using HA,?
Write the Henderson-Hasselbalch equation for a solution of propanoic acid (CH3CH2CO2H, pKa = 4.874) using HA, A– and the pKa value given in the expression.
pH=
Using this equation, calculate the quotient [A–]🇧🇷[HA] to
A) pH 4.44
B) pH 4.874
C) pH 5.20.
registration
I hate to be that guy, but I had a hard time reading what Kajola wrote. All credit goes to him, all I did was clean up the extra characters:
The Henderson-Hasselbalch equation is usually:
pH = pKa + log ([A−]🇧🇷[HA]🇧🇷
In this case it is:
4,874 + record ([A−]🇧🇷[HA]🇧🇷
A) 10^(4.44 – 4.874) = 0.368
= ~0.34
B) 10^(4.874 – 4.874) = 1
= 1
C) 10^(5.20 – 4.874) = 2.118
= ~2.1
<1> pH = pKa + log
🇧🇷[CH3CH2CO2-
]🇧🇷
[CH3CH2CO2H]
🇧🇷
pH = 4.874 +
registration
🇧🇷[CH3CH2CO2-
]🇧🇷
[CH3CH2CO2H]
🇧🇷
4.44 = 4.874 +
registration
🇧🇷[CH3CH2CO2-
]🇧🇷
[CH3CH2CO2H]
🇧🇷
[CH3CH2CO2-]
🇧🇷
[CH3CH2CO2H]
= 10^(4.44 –
4,874)
[CH3CH2CO2-]
🇧🇷
[CH3CH2CO2H]
= 0.368.
<2> 4.874 = 4.874
registration
🇧🇷[CH3CH2CO2-
]🇧🇷
[CH3CH2CO2H]
🇧🇷
[CH3CH2CO2-]
🇧🇷
[CH3CH2CO2H]
= 10^(4.874 –
4,874)
[CH3CH2CO2-]
🇧🇷
[CH3CH2CO2H]
= 1 .
<3> 5.20 = 4.874 +
registration
🇧🇷[CH3CH2CO2-
]🇧🇷
[CH3CH2CO2H]
🇧🇷
[CH3CH2CO2-]
🇧🇷
[CH3CH2CO2H]
= 10^(5.20 –
4,874)
[CH3CH2CO2-]
🇧🇷
[CH3CH2CO2H]
= 2.12 .Warning
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