Question

How many milliliters of 0.35 N sulfuric acid are necessary to react with 30.2 m/ of 0.52 N sodium hydroxide)

Answer 1

$z(H_2SO_4)=2$

$c(H_2SO_4)=0.35/z=0.35/2=0.175 M$

$z(NaOH)=1$

$c(NaOH)=0.52/1=0.52 M$

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

$n(NaOH)=0.52x0 0302=0.015704 mole$

$n(H_2SO_4)=n(NaOH)/2=0.015704/2=0.007852 mole$

$V=n(H_2SO_4)x1000/c(H_2SO_4)=0.007852x1000/0.175=44.9 ml$