If 50 mL of 0.2 M KOH is added to 40 mL o | vedclass

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(D) The reaction is: $HCOOH + KOH ightarrow HCOOK + H_2O$Initial moles of $HCOOH = 0.5 	imes 0.040 = 0.02 \ mol$Initial moles of $KOH = 0.2 	imes

Vedclass · Accepted Answer

$3.75$

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(D) The reaction is: $HCOOH + KOH ightarrow HCOOK + H_2O$Initial moles of $HCOOH = 0.5 	imes 0.040 = 0.02 \ mol$Initial moles of $KOH = 0.2 	imes 0.050 = 0.01 \ mol$After reaction,moles of $HCOOK$ (salt) $= 0.01 \ mol$Remaining moles of $HCOOH$ (acid) $= 0.02 - 0.01 = 0.01 \ mol$Using Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$$pK_a = -\log(1.8 	imes 10^{-4}) = 4 - \log(1.8) = 4 - 0.255 = 3.745$Since $[Salt] = [Acid] = 0.01 \ mol$,the ratio is $1$,and $\log(1) = 0$Therefore,$pH = pK_a = 3.745 \approx 3.75$

If $50 \ mL$ of $0.2 \ M \ KOH$ is added to $40 \ mL$ of $0.5 \ M \ HCOOH,$ the $pH$ of the resulting solution is $(K_a = 1.8 \times 10^{-4})$

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