agghh. Another chem problem

[JohnDoeDDS]

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Okay guys I keep on getting confused when I Get these type of problems!

How many ml of .05M HCl are required to turn 20ml of .05 KOH into a solution of ph 2.00?????


Okay I know that since the concentrations are both equal you will need the same amount of HCl to get a ph=7. So thats 20ml of HCl. Now I cant seen to figure out how to calculate how much ADDITIONAL HCl I will need to get the pH down.

 

[Manyak222]

Ok, so 20 ml and you have a bunch of water. Exactly 40 ml of it. You now need to add enough acid to get the concentration of H to .01 (eg 10^-2 moles) since ph=neg log H

scratch what i previously put. 1000 ml of HCl contains .05 moles, so you need this amount twice. 2000 plus 20 is 2020. I hope this is right

Okay guys I keep on getting confused when I Get these type of problems!

How many ml of .05M HCl are required to turn 20ml of .05 KOH into a solution of ph 2.00?????


Okay I know that since the concentrations are both equal you will need the same amount of HCl to get a ph=7. So thats 20ml of HCl. Now I cant seen to figure out how to calculate how much ADDITIONAL HCl I will need to get the pH down.

 

[JohnDoeDDS]

Still confused, Sorry. Answer is 40ml.

 

[GRAD]

Still confused, Sorry. Answer is 40ml.

Ok i think this:
It takes 20 ml to neutralize, and then i added M1V1=M2V2 and got another 20 ml, so add them together, and you get 40

 

[lifeisgood]

M1V1 = M2V2

M1 = 0.05 M
V1 = X

M2 = 0.01 (because [H+] = 0.01 gives pH of 2)
V2 = 40ml + X (because you want this molarity of 0.01 in the total volume)

(0.05)X = (0.01)(40+X)

(0.05)/(0.01) = (40+X)/X (use simple algebra)

5 = 40/X + 1

4 = 40/X

X = 40/4 = 10

Total volume added is 20 ml (to neutralize)+ 10 ml to bring to pH 2 = 30ml

Got it! 👍

 

[lifeisgood]

Still confused, Sorry. Answer is 40ml.

Just had this problem on Achiever #2. Definitely 30 ml.

 

[GRAD]

hes right..