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Oct 17, 2000
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I'm wondering if some of these are wrong in the book, or am I not getting it....:confused: I'll post this one question....if I find the rest, I'll post them also.

1) if 10 mL of 1M NaOH is titrated w/1M HCl to a pH of 2, what volume of HCl was added?

the answer says: 10.2mL added. First, add enough HCl to neutralize the sol'n. Since both acid and base are 1M, 10mL of HCl will neutralize 10mL of NaOH, from NaVa=NbVb. This produces 20mL of 0.5 NaCl sol'n.

Next calculate how much more HCl must be added to produce a [H+] of 1x10^-2. Let x be the amount of HCl to be added. The total volume of the solution will be (20+x) mL. Since this is now a dilution problem, the amount of HCl to be added can be found by using the formula: M1V1=M2V2....plugging in (1M)(xmL)=(0.01M)[(20+x)mL]

******* I don't know where they get 0.01M???******

continuing......When this equation is solved, x is found to have the value of 0.2. The final volume is 20.2mL so 10.2mL of HCl was added to the original NaOH sol'n.

Either I need to really put more time into gen chem, or this problem is tough! Where is the 0.01M from? Why is it not 0.5M? Is it a mistake?


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Sep 16, 2001
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i believe that the 0.01M was obtained from the concentration of H+ ions you need in solution to get a pH = 2. remember that
pH = -log [H+], so when you solve this, you get [H+] = 10^-2 which is equal to 0.01.
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