Question

What volume of 0.614m hal is required to titrate 18.8 ml of .619 m naoh

Answer 1

To titrate 18.8 mL of 0.619 M NaOH, you need approximately 18.9 mL of 0.614 M HCl.

Step-by-step explanation:

To determine the volume of 0.614 M HCl required to titrate 18.8 mL of 0.619 M NaOH, you can use the equation:

HCl + NaOH -> NaCl + H2O

First, calculate the number of moles of NaOH using the given concentration and volume:

1. 
   
3. 0.619 M NaOH * 0.0188 L = 0.0116 moles NaOH

Next, use the balanced equation to determine the mole ratio between HCl and NaOH:

1. 
   
3. 1 mole HCl / 1 mole NaOH

Finally, calculate the volume of 0.614 M HCl needed to react with the moles of NaOH:

1. 
   
3. 0.0116 moles NaOH * (1 mole HCl / 1 mole NaOH) = 0.0116 moles HCl
4. 
   
6. 0.0116 moles HCl / 0.614 M HCl = 0.0189 L or 18.9 mL