Question

It takes 83mL of a 0.45M NaOH solution to neutralize 235mL of an HCl solution.

Answer 1

Answer:

$M_{acid}=0.16M$

Explanation:

Hello there!

In this case, according to the reaction between hydrochloric acid and sodium hydroxide:

$HCl+NaOH\rightarrow NaCl+H_2O$

Whereas the 1:1 mole ratio of the acid to base allows us to write:

$n_{acid}=n_{base}$

At the equivalence point (complete neutralization); it is possible for us to write it in terms of molarities and volumes as follows:

$M_{acid}V_{acid}=M_{base}V_{base}$

In such a way, we solve for the molarity of the HCl to obtain:

$M_{acid}=\frac{M_{base}V_{base}}{V_{acid}} \\\\M_{acid}=\frac{83mL*0.45M}{235mL}\\\\M_{acid}=0.16M$

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