# 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 given pKa value in the expression.

RE:
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 given pKa value in the expression.
PH=
Using this equation, calculate the quotient [A–]/[HA] at
A) pH 4.44
B) pH 4.874
C) pH 5.20.

log

I hate to be that guy but I had a super hard time trying to read what Kajola wrote. All credit goes to him, all I did was clean up the extra characters:
Henderson-Hasselbalch equation is normally:
pH= pKa + log ([A−]/[HA])
In this case it’s:
4.874 + log ([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 +
log
{[CH3CH2CO2-
]/
[CH3CH2CO2H]
}
4.44 = 4.874 +
log
{[CH3CH2CO2-
]/
[CH3CH2CO2H]
}
[CH3CH2CO2-]
/
[CH3CH2CO2H]
= 10^(4.44 –
4.874)
[CH3CH2CO2-]
/
[CH3CH2CO2H]
= 0.368 .
<2> 4.874 = 4.874

Also Read :   come si dice ti voglio bene in greco?

log

{[CH3CH2CO2-
]/
[CH3CH2CO2H]
}
[CH3CH2CO2-]
/
[CH3CH2CO2H]
= 10^(4.874 –
4.874)
[CH3CH2CO2-]
/
[CH3CH2CO2H]
= 1 .
<3> 5.20 = 4.874 +
log
{[CH3CH2CO2-
]/
[CH3CH2CO2H]
}
[CH3CH2CO2-]
/
[CH3CH2CO2H]
= 10^(5.20 –
4.874)
[CH3CH2CO2-]
/
[CH3CH2CO2H]
= 2.12 .Take care