Question

What volume of a 3.5 M HCL is required to completely neutralize 50.0 ml of a 2.0 M NaOH

Answer 1

NaOH:
V=50mL=0,05L
Cm = 2M

n = Cm*V = 0,05L * 2M = 0,1mol

HCl + NaOH ⇒ NaCl + H₂O
1mol : 1mol
0,1mol : 0,1mol

HCl:

n = 0,1mol
Cm = 3,5M

n 0,1mol
V=---------- = ----------- ≈ 0,029dm³ = 29mL of 3,5M HCl
Cm 3,5 mol/dm³

Answer 2

Answer: The volume of HCl required will be 28.57 mL.

Step-by-step explanation:

To calculate the volume of acid, we use the equation given by neutralization reaction:

`n_1M_1V_1=n_2M_2V_2`

where,

`n_1,M_1\text{ and }V_1` are the n-factor, molarity and volume of acid which is HCl

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is NaOH

We are given:

`n_1=1\\M_1=3.5M\\V_1=?mL\\n_2=1\\M_2=2M\\V_2=50mL`

Putting values in above equation, we get:

`1* 3.5* V_1=1* 2* 50\\\\V_1=28.57mL`

Hence, the volume of HCl required will be 28.57 mL.