What is the pH of a solution prepared by mixi | vedclass

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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

$3.57$

Vedclass · Answer

(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$.

Buffer Solution	Volume of $0.1 \ M$ Weak Acid (mL)	Volume of $0.1 \ M$ Sodium Salt (mL)
$I$	$4.0$	$4.0$
$II$	$4.0$	$40.0$
$III$	$40.0$	$4.0$
$IV$	$0.1$	$10.0$

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 $HCOOH = 3.75$):-

Similar Questions

Which statement is correct for a mixture of $1 \ M \ NaCl$ and $1 \ M \ HCl$?

An example of a neutral buffer is:

Buffer Solution Volume of $0.1 \ M$ Weak Acid (mL) Volume of $0.1 \ M$ Sodium Salt (mL)

$I$ $4.0$ $4.0$

$II$ $4.0$ $40.0$

$III$ $40.0$ $4.0$

$IV$ $0.1$ $10.0$

Which of the two sets of buffer solutions have the least $pH$?

Which one of the following is not a buffer solution?

$A$ solution is prepared by mixing $0.01 \ mol$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$ in $100 \ mL$ of water. The $pH$ of the resulting solution is. . . . . . .
[Given : $pK_{a1}$ and $pK_{a2}$ of $H_2CO_3$ are $6.37$ and $10.32$,respectively; $\log 2=0.30$ ]

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