A25-ml sample of 0.160m solution of naoh is titrated with 17 ml of an unknown solution of
h2
so4
. what is the molarity of the sulfuric acid solution?
a. 0.235m h2
so4
b. 0.117m h2so4
c. 0.004m h2
so4
d. 0.002m h2so4

V ( NaOH ) = 25 mL in liters : 25 / 1000 => 0.025 L

M ( NaOH ) = 0.160 M

V ( H2SO4) = 17 mL / 1000 => 0.017 L

M ( H2SO4) = ?

number of moles NaOH:

n = M x V = 0.160 x 0.025 => 0.004 moles NaOH

Mole ratio :

2 NaOH + H2SO4 = Na2SO4 + 2 H2O

2 moles NaOH 1 mole H2SO4
0.004 moles NaOH ? moles H2SO4

moles H2SO4 = 0.004 x 1 / 2

= 0.002 moles of H2SO4

M ( H2SO4) = n / V

M = 0.002 / 0.017

 = 0.117 M H2SO4

Answer B

hope this helps!

Well to my expert knowledge and my exquisite calculations i think the the negative ion # 412 is the key to sucess