Question

Calculate volume of 2 molar sodiumhydroxide required for complete neutralization of 200ml of 3 molar sulphuric acid solution?

Answer 1

c(NaOH) = 2mol/L
V(H2SO4) =200mL = 0.2 L
c(H2SO4) = 3 mol/L
V(NaOH) - ?
2NaOH (aq) + H2SO4(aq) = Na2SO4 (aq) + 2H2O (l) (1)
Find amount of chemical substance of H2SO4 in 0.2 L of solution:
c=n/V  n= cV;
n(H2SO4 )= 30.2 = 0.6 (mol).
According to equation (1) 2 mol of NaOH react with 1mole of H2SO4
We have x mol of NaOH that react with 0.6 mol of H2SO4.
Solve the proportion:
2/x= 1/0.6
x=1.2 mol
n(NaOH) = 1.2 mol
Find volume of NaOH solution:
c=n/V  V=n/c;
V (NaOH) = 1.2/2 = 0.6 (L) = 600 (mL).