Question

It take 38.70cm³ of 1.90m NaoH to neutralize 10.30cm³ of H2so4 in a battery, calculate the molar concentration of H2so4

Answer 1

Answer:

$M_{acid}=3.57M$

Explanation:

Hello there!

In this case, since this acid-base neutralization is performed in a 1:2 mole ratio of acid to base as the former is a diprotic acid (two hydrogen ions in the molecule), we can write the following equation:

$2M_{acid}V_{acid}=M_{base}V_{base}$

In such a way, we can solve for the molarity of the acid, given the molarity and concentration of the NaOH base and the volume of the acid:

$M_{acid}=\frac{M_{base}V_{base}}{2V_{acid}}$

Thus, we plug in the given data to obtain:

$M_{acid}=\frac{38.70cm^3*1.90M}{2(10.30cm^3)} \\\\M_{acid}=3.57M$

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