Question

How many mL of 0.150M NaOH (base) solution are required to neutralize 35.0mL of 0.220M HCl (acid) solution?

Answer 1

`V_B=51.3mL`

Step-by-step explanation:

Hello,

In this case, one uses the titration equation that comes from the moles equivalence during the neutralization:

`M_AV_A=M_BV_B\\V_B=(M_AV_A)/(M_B) \\V_B=(0.220M*35.0mL)/(0.150M) \\V_B=51.3mL`

Such equality is used since the acid is monoprotic and the base has just one hydroxile allowing the moles to be equal at the equivalence point.

Best regards.

Answer 2

The neutralization reaction between an acid and a base is given through the equation,
M₁V₁ = M₂V₂
Substituting the known values from the given,
(0.150 M)(V₁) = (35 mL)(0.220M)
The value of V₁ from the equation above is 51.33 mL.

Thus, 51.33 mL of NaOH is needed for the process.