calculate volume (in ml) of 0.300M NaOH required to neutralize 20.0ml of 0.250M H2SO4?

Question

please help

Answer 1

The balance of the equation is

2NaOH + H2SO4 ==> 2H2O + Na2SO4

therefore you will need twice as many molecules of NaOH to neutralize a given number of molecules of H2SO4

0.250M = 0.250moles/l or 0.250mmoles/ml

in 20ml of a 0.250M solution of H2SO4 there are
0.250x20 mmoles = 5mmoles

So you´ll need 10mmoles NaOH
your solution of NaOH is 0.3M = 0.3mmoles/ml
So you will find 10mmoles in:
10/0.3 = 33.33ml

answer: 33.33ml of NaOH 0.3M
hope it is clear

Answer 2

This is a normality problem since sulfuric acid is di-basic. Thus you will need twice as much NaOH to neutralize it as would need for a monobasic acid. So, figure the moles of H2SO4= 20mLx.250M= 5 millimoles. Now double it. Then 10millimoles= 0.3 V, V = 33.3 mL

Answer 3

2NaOH + H2SO4 => Na2SO4 + 2H2O

=> 2 mols of NaOH neutralises 1 mol H2SO4

no of mols of H2SO4 in 20.0 ml 0.250M H2SO4 = 0.25 x 20/1000 = 0.005 mol.

no of mols of NaOH reqd to neutralize 0.005 mol H2SO4 = 0.005 x 2 = 0.01 mol

in 0.300M, 1000 ml soln contains 0.3 mol NaOH

hence, 0.01 mol of NaOH is contained in = 1000 x 0.01/0.3=33.33 ml

so. reqd vol of soln = 33.33 ml