[Blonde]

How many milliliters of 0.50N HCl are needed to neutralize 40 mL of 0.80N NaOH?

[mookxi]

How many milliliters of 0.50N HCl are needed to neutralize 40 mL of 0.80N NaOH?

What I did was first write an equation.

NaOH + HCl --> NaCl + H₂O

then I found the mole of NaOH by going n = c x v = 0.032

there is a ratio of 1:1 so n(NaCl) = n(HCl) = 0.032

Volume = n / c = 0.032/0.5 = 0.064L

that's the answer.. i should think.

[Donaldson Tan]

Ref: http://www.chemicalforums.com/index.php?topic=9037.0
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HCl + NaOH -> NaCl + H₂O

The stoichiometric ratio for HCl: NaOH is 1:1, so
volume of HCl required = (1/1)  * (40*0.80) / (0.50) = 64ml = 0.064L

[Borek]

64ml = 0.0064L

[Donaldson Tan]

corrected

[kayamusty]

n.v=n.v
0.5xv=40x0.8

v=64 ml