Question

a 20ml sample of hcl was titrated with the 0.0220 M Naoh. to reach the endpoint required 23.72 mL of the NaOh. Calculate the molarity of HCL

Answer 1

Answer:

Molarity of HCl=0.026092M

Explanation:

The equation for the reaction is;

HCl + NaOH ⇒ NaCl + H2O

Using the formular, $\frac{C_{A}V_{A}}{C_{B}V_{B} }=\frac{nA}{nB}$    ..........equ1

where$C_{A}$ is the concentration of Acid,

          $V_{A}$ is volume of acid

          $C_{B}$ is concentration of the base

          $V_{B}$ is volume of the base

          nA is the number of moles of Acid

          nB is number of moles of base

nA = 1,    nB=1 , $V_{A}=20ml$, $C_{B}=0.022M$, $V_{B}=23.72mL$

we will input these values into equation1 to solve for $C_{A}$

$\frac{C_{A}*20}{0.022*23.72}=\frac{1}{1}$

$C_{A}*20=0.022*23.72$

$C_{A}=0.026092M$