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

`n_(base)=3.90x10^(-3)molNaOH`

Step-by-step 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(mol)/(L) \\\\n_(base)=3.90x10^(-3)molNaOH`

Best regards!