Question
In: Chemistry
In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of...
In the titration of 25.00 mL of 0.100 M acetic acid, what is the pH of the resultant solution after the addition of 15.06 mL of 0.100 M of KOH? Assume that the volumes of the solutions are additive. Ka = 1.8x10-5 for acetic acid, CH3COOH.
Solutions
Expert Solution
no of moles of CH3COOH = molarity * volume in L
= 0.1*0.025 = 0.0025 moles
no of moles of KOH = molarity * volume in L
= 0.1*0.01506 = 0.001506 moles
CH3COOH + KOH -----------------> CH3COOK + H2O
I 0.0025 0.001506 0
C -0.001506 - 0.001506 0.001506
E 0.000994 0 0.001506
PKa = -logKa
= -log1.8*10^-5
= 4.75
PH = PKa + log[CH3COOK]/[CH3COOH]
= 4.75 + log0.001506/0.000994
= 4.75 + 0.1804
= 4.9304 >>>>>>>>>>>>>>answer
queen_honey_blossom answered 2 years agoRelated Solutions
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