Neutralization reaction?

Question

If 26.24 mL of a standard 0.1650 M NaOH solution is required to neutralize 33.03 mL of H2SO4, what is the molarity of the acid solution?

Please explain your answer step by step. Thanks!

Answer 1

26.24 ml = 0.02624 L
0.02624L * 0.1650 mols/L = 4.33*10^-3 mols used to neutralize
so there were 4.33*10^-3 mols of H+
H2SO4, HAS 2 mols of H+, so divide this number by 2
4.33*10^-3/2 = 2.16*10^-3 = 0.0021648 mols of H2SO4.

0.0021648 mols H2SO4 / 0.03303L= 0.0655 M of H2SO4

Answer 2

First write down the reaction equation:-
2NaOH + H2SO4 = Na2SO4 + 2H2O
Notice the molar ratios in the equation are:- 2:1::1:2
Using the general 'moles' equation :

moles = [Conc/M] x volume(cm^3)/ 1000(cm^3)

NB the '1000cm^3 is required in the calculation to convert 'M' moles per litre to 'moles per cm^3'.

moles(NaOH) = 0.1650 x 26.24 /1000 = 0.0043296 = 4.33 x 10^-3mol.
Moles(NaOH) = 4.33 x 10^-3
Therefore by molar ratios above 2 moles of NaOH reacts with 1 mole H2SO4, so the moles value calculated for NaOH is divided by 2 to obtain the moles value for Sulphuric Acid.
Moles (H2SO4) = 4.33 x 10^-3 /2 = 2.16 x 10^-3 mol

Going back to the general 'moles' equation and algebraically rearranging.

moles = [conc] x volume / 1000cm^3
[conc] = moles x 1000cm^3 / volume
[conc] = 2.16 x 10^-3 x 1000 / 33.03 = 0.06539 moles per litre
[conc] = 6.54 x 10^-2 M

Hope this helps !!!!

Answer 3

2NaOH + H2SO4 --> Na2SO4 + 2H2O
mL x M = millimoles
26.24 x 0.1650 = 4.3296 millimoles of NaOH
Since 2 moles of NaOH are required to neutralize 1 mole of H2SO4, you started with 2.1648 millimoles of H2SO4 (4.3296/2)

2.1648 millimoles = 33.03 mL x M
M=0.0655
H2SO4 was 0.0655 Molar