Achiever #2 Question 41 GC

[bittersweet008x]

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How many mL of 0.05 M HCL are required to turn 20 mL of 0.05 M KOH into a solution of pH 2.00?

The answer is 10.

Achiever gives a solution and my response to it is just HUH?

any help would be greatly appreciated!🙂

 

[loganheinrich]

Are you sure that's the answer? It doesn't compute in my mind... if you add half of the amount of strong acid to strong base, you should be left with half water and the rest strong base, right?

To have a pH of 2, you would need 10^2 H+ ions, so logically you'd need to consume all of the strong base with 20 mL of HCl and then add HCl to the point of having a H+ concentration of 10^2, maybe using m1v1=m2v2

 

[bittersweet008x]

so sorry.. the answer is 30..

 

[jefff]

i got 28 mL...

20mL of both *since strong acid and strong base* will neutralize to pH 7 (total solution volume being 40mL)

so you want to know how much more acid is then needed to decrease the pH further to 2 (total [H+] = 0.01 M*for new solution*)


M1V1 = M2V2
(5e-2 M)(X mL) = (1e-2 M)(4e1 mL + X mL)

(5e-2)(X) = (4e-1 + 1e-2X)

(X) - (1e-2X) = (4e-1)/(5e-2)
*since this is like 1-.01 which would be .99...im just going keep it at 1 for simplicity*

X = .8 e1
X = 8 mL

therefore original volume of 20 mL HCL (needed to neutralize the buffer) + 8 mL (to bring to pH 2)

total HCL volume needed = 28 mL