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

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

Expert Solution

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

HCOOH + NaOH --------------------- HCOONa + H2O

Initial 0.0025 0.00239 0

change - 0.00239 - 0.00239 + 0.00239

equilibrium 0.00011 0 + 0.00239

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

PH = 3.79 + log(0.00239/0.00011)

PH = 5.13

PH of the solution = 5.13

queen_honey_blossom answered 10 months ago

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