What would the pH be for a titration of 40mL of 0.1M formic acid with 40 mL (and

Question

What would the pH be for a titration of 40mL of 0.1M formic acid with 40 mL (and 20mL) of NaOH? Ka=1.8x10^-4. what would the curve look like? At what pH would [HCOOH]=[HCOO-]?

Explanation / Answer

1. When NaOH = 40ml; Moles HCOOH=40.0mL x 0.100M=4.00mmol; Moles OH-=40mL x 0.100M=4mmol; So, [HCOO-]= 4mmol/(40+40)ml= 0.05 M; Now COOH- will react with water to form HCOOH; COOH- + H2O-> HCOOH+ OH-; Assume x moles of HCOOH and OH- produces and 0.05-x moles of COOH- remain then Kb=x^2/(0.05-x) where Kb=Kw/Ka= 5.55*10^-11; Solve for x and get x as x=1.66*10^-6; So, pOH= -log[x]= 5.77; Therefore, pH= 14-pOH= 8.22; When NaOH =20ml; Moles HCOOH=40.0mL x 0.100M=4.00mmol; Moles OH-=40mL x 0.100M=2mmol; So,[HCOO-]= 2mmol; and [HCOOH]=2mmol; so, pH=pKa+log [Base]/[Acid]; putting in values pH= 3.74; 2. at pH=pKa, [HCOOH]=[HCOO-], so pH= 3.74

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