Question

In a titration experiment, 31. 4 ml of 1. 120 m hcooh is neutralized by 16. 3 ml of ba(oh)2. what is the concentration of the ba(oh)2 solution?

Answer 1

In a titration experiment, 31. 4 ml of 1. 120 M HCOOH is neutralized by 16. 3 ml of $Ba(OH)_{2}$. So, the concentration of the $Ba(OH)_{2}$  solution is 23.16 M

Calculation,

The formula for the dilution of a solution is given as:

$M_{1} V_{1}$ = $M_{2} V_{2}$

Where M is molarity and V is the volume of the solution in liters ( L ).

Given data,

$M_{1}$ =  120  M

$V_{1}$ =31. 4 ml

$M_{2}$ = ?

$V_{2}$ = 16. 3 ml

120  M ×31. 4 ml  =  $M_{2}$  ×  16. 3 ml

$M_{2}$  = 120  M ×31. 4 ml / 16. 3 ml =23.16 M

In a titration experiment, the concentration of the $Ba(OH)_{2}$  solution is 23.16 M

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