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45.0 mL of a Naoh solution is titrated with 0.12 M HCl. If 29.6 mL is required for neutralization,
what is the concentration of the NaOH solution?

Respuesta :

SvetkaChem

Answer:

M(NaOH) = 0.0789 M

Explanation:

NaOH + HCl ---> NaCl + H2O

M(NaOH)*V(NaOH) =M(HCl)*V(HCl)

M(NaOH) *45.0 mL (NaOH) = 0.12 M(HCl) * 29.6 mL(HCl)

M(NaOH) = 0.12*29.6/45.0  = 0.0789 M

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