Methanoic acid has a Ka=1.6X10-4. Calculate the pH of the final solution when 23.90 mL of...

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Methanoic acid has a Ka=1.6X10-4. Calculate the pH of the finalsolution when 23.90 mL of 0.100M NaOH is added to 25.00 mL of 0.100M methanoic acid

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3.8 Ratings (442 Votes)

Ka = 1.6 x10^-4

NaOH = 23.90 ml of 0.100M

number of moles of NaOH = 0.100M x 0.02390L = 0.00239 moles

Methanoic acid = 25.00ml of 0.100M

number of moles of Methanoic acid = 0.100M x 0.02500L = 0.0025 moles

Ka = 1.6 x 10^-4

-log(Ka) = -log(1.6 x10^-4)

Pka = 3.79

Methanoic acid = HCOOH

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â HCOOHÂ Â Â Â +Â Â Â NaOH ---------------------Â Â HCOONaÂ Â Â Â Â + H2O

InitialÂ Â Â Â Â Â Â Â Â Â 0.0025Â Â Â Â Â Â Â Â Â Â 0.00239Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

change Â Â - 0.00239Â Â Â Â Â Â Â - 0.00239Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â + 0.00239

equilibriumÂ Â Â Â Â Â Â Â Â Â Â Â Â Â 0.00011Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â + 0.00239

PH = Pka + log[salt]/[acid]

PH = 3.79 + log(0.00239/0.00011)

PH = 5.13

PH of the solution = 5.13

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