GC neutralization reactions help! plz

[orthdent786]

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How much 2 M H2SO4 is required to neutralize 200 ml of 2M NaOH?
So

NaVa=NbVb
Solve for Va= Nb Vb/ Na

(200) (2)/ 2 = 200ml

BUT that is the wrong answer this is from Kaplan Subject Test 5 they say it should be

(200) (2) / 4 --> where do they get the 4 from?

thanks in advance!!

 

[wigglytooth]

How much 2 M H2SO4 is required to neutralize 200 ml of 2M NaOH?
So

NaVa=NbVb
Solve for Va= Nb Vb/ Na

(200) (2)/ 2 = 200ml

BUT that is the wrong answer this is from Kaplan Subject Test 5 they say it should be

(200) (2) / 4 --> where do they get the 4 from?

thanks in advance!!

The 4 is from multiplying the concentration of H2SO4 by 2, since you have two equivalents of H+. Normality (N) is actually equivalent moles/L and not just mol/L (which is Molarity, M).

 

[How Sleepy??!!]

Yeh...just remember that normality is always double the molarity when the acid has 2 H's. And it works the other way too, when solving for molarity given the normality (but i didnt see that on my DAT).

 

[nate28]

How much 2 M H2SO4 is required to neutralize 200 ml of 2M NaOH?
So

NaVa=NbVb
Solve for Va= Nb Vb/ Na

(200) (2)/ 2 = 200ml

BUT that is the wrong answer this is from Kaplan Subject Test 5 they say it should be

(200) (2) / 4 --> where do they get the 4 from?

thanks in advance!!

I'm sorry, I don't like to work in N.

So 0.2 L of 2M NaOH = 0.4 mols OH-
For every 1 mol OH-, you need 1/2 as much H2SO4 b/c each H2SO4 can donate 2 H+... so 0.5 mols
That means you need 0.2 mols of H2SO4.
2 M = 0.2 mol / L
L = 0.1 L = 100 mL

Ok: (0.2 L)(2 M NaOH) [(1 mol H2SO4)/(2 mol NaOH)] = 0.2 mol H2SO4 / XL = 2 M H2SO4
Solve for XL: 0.2 mol H2SO4 / 2 M H2SO4 = 0.1 L

(0.2 L )(2 M )/(2 mol)(2 M)
0.4/4 = 0.1