Question

you are given a solution of hcooh (formic acid) with an approximate concentration of 0.20 m and you will titrate this with a 0.1105 m naoh. if you add 20.00 ml of hcooh to the beaker before titrating, approximately what volume of naoh will be required to reach the end point? view available hint(s) 20.0 ml 72.4 ml 11.1 ml 36.2 ml submit part b you are given a solution of hcooh (formic acid) with an approximate concentration of 0.20 m and you will titrate this with a 0.1105 m naoh. you add 20.00 ml of hcooh to the beaker before titrating, and it requires 35.43 ml of naoh to reach the end point. what is the concentration of the hcooh solution? view available hint(s) 0.0624 m 0.1105 m 0.2000 m 0.1958 m

Explanation:

Step1: Recall the titration formula

$n_1V_1 = n_2V_2$ for a monoprotic - monobasic titration, where $n$ is the number of moles and $V$ is the volume. For formic acid ($HCOOH$) and sodium hydroxide ($NaOH$), the mole - ratio is 1:1. So $M_1V_1 = M_2V_2$, where $M$ is the molarity.

Part A

Step1: Identify the given values

$M_1$ (molarity of $HCOOH$) = 0.20 M, $V_1$ (volume of $HCOOH$) = 20.00 mL, $M_2$ (molarity of $NaOH$) = 0.1105 M. We need to find $V_2$ (volume of $NaOH$).
Using the formula $M_1V_1 = M_2V_2$, we can solve for $V_2$:
$V_2=\frac{M_1V_1}{M_2}$

Step2: Substitute the values

$V_2=\frac{0.20\ M\times20.00\ mL}{0.1105\ M}=\frac{4.00\ mmol}{0.1105\ M}\approx36.2\ mL$

Part B

Step1: Identify the given values

$V_1$ (volume of $HCOOH$) = 20.00 mL, $M_2$ (molarity of $NaOH$) = 0.1105 M, $V_2$ (volume of $NaOH$) = 35.43 mL. We need to find $M_1$ (molarity of $HCOOH$).
From the formula $M_1V_1 = M_2V_2$, we can solve for $M_1$:
$M_1=\frac{M_2V_2}{V_1}$

Step2: Substitute the values

$M_1=\frac{0.1105\ M\times35.43\ mL}{20.00\ mL}=\frac{3.915\ mmol}{20.00\ mL}= 0.1958\ M$

Answer:

Part A

D. 36.2 mL

Part B

D. 0.1958 M