Answer:
pH = 12.33
Explanation:
Lets call HA = butanoic acid and A⁻ butanoic acid and its conjugate base butanoate respectively.
The titration reaction is
HA + KOH ---------------------------- A⁻ + H₂O + K⁺
number of moles of HA : 118.3 ml/1000ml/L x 0.3500 mol/L = 0.041 mol HA
number of moles of OH : 115.4 mL/1000ml/L x 0.400 mol/L = 0.046 mol A⁻
therefore the weak acid will be completely consumed and what we have is the unreacted strong base KOH which will drive the pH of the solution since the contribution of the conjugate base is negligible.
n unreacted KOH = 0.046 - 0.041 = 0.005 mol KOH
pOH = - log (KOH)
M KOH = 0.005 mol / (0.118.3 +0.1154)L = 0.0021 M
pOH = - log (0.0021) = 1.66
pH = 14 - 1.96 = 12.33
Note: It is a mistake to ask for the pH of the acid solution since as the above calculation shows we have a basic solution the moment all the acid has been consumed.
The pH of the acid solution is mathematically given as
pH= 12.33
Question Parameter(s):
An analytical chemist is titrating 118.3 mL of a 0.3500 M solution of butanoic acid (HC3H7CO2)
a 0.400 M solution of KOH
pKa of butanoic acid is 4.82.
Generally, the equation for the Chemical Reaction is mathematically given as
HA + KOH ⇄ A⁻ + H₂O + K⁺
Therefore
pOH = - log (KOH)
M of KOH = 0.005 mol / (0.118.3 +0.1154)
M of KOH = 0.0021 M
Hence
pOH = - log (0.0021)
pOH= 1.66
In conclusion, pOH= 1.66
pH = 14 - 1.96
pH= 12.33
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