Kaplan Acid and Bases Question #5

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bigmann4208

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10+ Year Member
The question reads if 10 ml of 1M NaOH is titrated with 1M of HCl to a pH of 2, what volume of HCl was added?

I did N1V1=N2V2 but that got me 10.0 mL and the answer is 10.2 mL. Any help would be appreciated.
 
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You have to take the pH into account. I tried to figure it out but got nowhere. I guess my gen chem skills are rusty...which is not a good sign. But I do know that [H+] = 10^-pH.....so I guess the final hydrogen ion concentration is .01 M in this case but then again I'm really not quite sure. Perhaps using one of those initial,change,final diagrams would be of help. Please post how you get the final product if you figure it out. Sorry this post wasn't much help.

-MCPD
 
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I don't know if this is the best way but the way i did is:

Final H+ conc. = 10^-2 = 0.01

When we mix NaOH and HCl , there conc. changes so the we calculate the new conc.

We should assume x for volume of HCl required.

New conc. of OH- = (10mL*1M)/10mL+xmL
New conc. of H+ = (xmL*1M)/xmL+10mL

OH- + H+ -----> H2O
S 10/(x+10) x/(x+10) -----
C -(10/x+10) -(10/x+10) -----
N 0 0.01 -----

(x/x+10) - (10/(x+10)) = 0.01

x = 10.2 which the vol of HCl req. to reach pH of 2
 
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