7. What volume of a 0.10 mol/l HCl solution is needed to neutralize 10 ml of a 0.15
mol/l LiOH solution (5 points)

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

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