An analytical chemist is titrating 118.3 ml of a 0.3500 m solution of butanoic acid (hc3h7co2) with a 0.400 m solution of koh. the pka of butanoic acid is 4.82. calculate the ph of the acid solution after the chemist has added 115.4 ml of the solution to it. note for advanced students: you may assume the final volume equals the initial volume of koh the solution plus the volume of solution added. round your answer to 2 decimal places.

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.

2b2t

explanation:

2b2t