Question

It takes 38 mL of 0.75 M NaOH solution to completely neutralize 155 mL of a sulfuricacid solution (H2SO4). a.Write a balanced equation for the neutralizationof NaOH with H2SO4

Answer 1

Answer:

a. $H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

b. 0.092 M

Explanation:

Hello!

In this case, since for this titration process we know the used volume of 0.75-M sodium hydroxide needed to neutralize 155 mL of the acid, we first need to write the undergoing chemical reaction between them:

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

This, since there is 1:2 mole ratio between the acid and the base, at the equivalent point we must respect:

$2*n_{acid}=n_{base}$

That in terms of concentrations and volumes is:

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

Thus, the concentration of acid was:

$M_{acid}=\frac{M_{base}V_{base}}{2*V_{acid}} \\\\M_{acid}=\frac{0.75M*38mL}{2*155mL} \\\\M_{acid}=0.092M$

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