Question

g How many moles of NaOH are present in a sample if it is titrated to its equivalence point with 44.02 mL of 0.0885 M H2SO4? 2NaOH(aq) + H2SO4(aq) --> Na2SO4(aq) + 2H2O(l)

Answer 1

Answer:

$n_{base}=3.90x10^{-3}molNaOH$

Explanation:

Hello!

In this case, since the sulfuric acid and sodium hydroxide react in a 1:2 mole ratio, given the reaction, we realize they have the following mole ratio at the equivalence point:

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

Which in terms of concentrations and volumes is:

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

Thus, we can plug in the volume and concentration of acid to find the moles of base:

$n_{base}=0.04402L*0.0885\frac{mol}{L} \\\\n_{base}=3.90x10^{-3}molNaOH$

Best regards!