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Shkiper50 [21]
3 years ago
6

If 26.35 ml of a standard 0.1650 m naoh solution is required to neutralize 35.00 ml of h2so4, what is the molarity of the acid s

olution?
Chemistry
2 answers:
vovikov84 [41]3 years ago
6 0
 <span>26.35 ml NaOH x (.1650 moles NaOH / 1000 ml ) x (1 mole H2SO4 / 2 moles NaOH) = moles H2SO4 = .002174 moles H2SO4 

molarity of H2SO4 = .002174 / .035 L = .06211 M</span>
o-na [289]3 years ago
3 0

Answer : The molarity of the H_2SO_4 is, 0.06211 M

Explanation :

Using neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

n_1 = basicity of an acid (H_2SO_4) = 2

n_2 = acidity of a base (NaOH) = 1

M_1 = concentration or molarity of H_2SO_4 = ?

M_2 = concentration of NaOH = 0.1650 M

V_1 = volume of H_2SO_4 = 35.00 mL

V_2 = volume of NaOH = 26.35 mL

Now put all the given values in the above law, we get the concentration of the H_2SO_4.

2\times M_1\times 35.00mL=1\times 0.1650M\times 26.35mL

M_1=0.06211M

Therefore, the molarity of the H_2SO_4 is, 0.06211 M

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