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(A) The reaction between sodium formate $(HCOONa)$ and hydrochloric acid $(HCl)$ is:$HCOONa + HCl \rightarrow HCOOH + NaCl$Initial moles:$n(HCOONa) =

Vedclass · Accepted Answer

$3.57$

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(A) The reaction between sodium formate $(HCOONa)$ and hydrochloric acid $(HCl)$ is:$HCOONa + HCl ightarrow HCOOH + NaCl$Initial moles:$n(HCOONa) = 0.250 \ M 	imes 20 \ mL = 5 \ mmol$$n(HCl) = 0.100 \ M 	imes 30 \ mL = 3 \ mmol$Since $HCl$ is the limiting reagent,it will be completely consumed.Final moles after reaction:$n(HCOONa) = 5 - 3 = 2 \ mmol$$n(HCOOH) = 3 \ mmol$Total volume of the solution = $20 \ mL + 30 \ mL = 50 \ mL$.The resulting solution is a buffer solution containing a weak acid $(HCOOH)$ and its conjugate base $(HCOO^-)$.Using the Henderson-Hasselbalch equation:$pH = pK_a + \log \frac{[Salt]}{[Acid]}$$pH = 3.75 + \log \frac{(2 \ mmol / 50 \ mL)}{(3 \ mmol / 50 \ mL)}$$pH = 3.75 + \log \frac{2}{3}$$pH = 3.75 + (0.301 - 0.477) = 3.75 - 0.176 = 3.574$Rounding to two decimal places,$pH = 3.57$.

What is the $pH$ of a solution prepared by mixing $20 \ mL$ of $0.250 \ M$ sodium formate with $30 \ mL$ of $0.100 \ M$ $HCl$ ($pK_a$ for $HCOOH = 3.75$):-

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