Question

How many mL of 25 M NaOH (sodium hydroxide) is required to react with 50.0 mL of 150 M H2SO4 (sulfuric acid). Round your answer to 3 decimal places.

Answer 1

Molarity$=$ $Moles of solute\over Liter of solution$

The chemical equation for the reaction is;

$NaOH+H_2SO_4\to Na_2SO_4+H_2O$

$50mL\times$ $150MH_2SO_4\over 1000mL$ $=7.5 mol$

$Molecular weight of NaOH=39.997g/mol$

$39.997g/mol NaOH\over 7.5molH_2SO_4$ $=5.33mL$

$NaOH=5.33mL$