Question

It took 14.50 mL of 0.455M NaOH to fully neutralize 12.0mL of HCl. What is the concentration of the HCl?

HCl + NaOH \rightarrow→ NaCl + H2O

Answer 1

0.550 M HCl

Step-by-step explanation:

M1V1 = M2V2

M1 = 0.455 M NaOH

V1 = 14.50 mL NaOH

M2 = ?

V2 = 12.0 mL HCl

Solve for M2 --> M2 = M1V1/V2

M2 = (0.455 M)(14.50 mL) / (12.0 mL) = 0.550 M HCl

Answer 2

The appropriate answer is "0.549 M".

Step-by-step explanation:

The given values are:

N₁ = 14.50 mL

V₁ = 0.455 M

N₂ = 12 mL

Let

V₂ = C = ?

As we know,

⇒
`N_1* V_1=N_2* V_2`

On substituting the values, we get

⇒
`14.50* 0.455 = 12* C`

⇒
`6.5975=12* C`

⇒
`C=(6.5975)/(12)`

⇒
`=0.549 \ M`