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

Respuesta :

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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. 
sebassandin

Answer:

[tex]V_B=51.3mL[/tex]

Explanation:

Hello,

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

[tex]M_AV_A=M_BV_B\\V_B=\frac{M_AV_A}{M_B} \\V_B=\frac{0.220M*35.0mL}{0.150M} \\V_B=51.3mL[/tex]

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.

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