Respuesta :
Answer: The volume of a 0.10 mol/l HCl solution needed to neutralize 10 ml of a 0.15 mol/l LiOH solution is 15 ml
Explanation:
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is [tex]HCl[/tex]
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is LiOH.
We are given:
[tex]n_1=1\\M_1=0.10mol/L\\V_1=?mL\\n_2=1\\M_2=0.15mol/L\\V_2=10mL[/tex]
Putting values in above equation, we get:
[tex]1\times 0.10\times V_1=1\times 0.15\times 10\\\\V_1=15mL[/tex]
Thus the volume of a 0.10 mol/l HCl solution needed to neutralize 10 ml of a 0.15 mol/l LiOH solution is 15 ml
