Dat Acheiver Test 2 Gen Chem Question 1 HELP!!!

[paradigm2200]

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Hey guys....reviewing my tests on DAT acheiver and this question completely stumps me....would love if you guys could explain it to me...

anyway...

The question is from test 2, gen chem question 1:

How many milliliters of 0.05M HCl are required to turn 20 ml of 0.05M KOH into a solution of pH 2.00?

The solution is the following (as given by DAT acheiver):

Correct Answer: E

It will be easier if you can break up the calculation into two parts and sum up the HCl volumes at the end of it.

The amount of HCl required for neutralization: 20 ml ---------- (1)

Volume of the neutralized solution: 20 + 20 = 40 ml ---------- (2)

Next, you will need to determine the additional amount of HCl, y ml, required to turn the neutralized solution into an acidic one with pH = 2.00.

Using the pH equation,

pH = -log [H+]

2 = -log [H+]

[H+] = 1 x 10-2 M ---------- (3)

Applying simple proportional relation,

[y(0.05)/1000] / [(40 + y)/1000] = 1 x 10-2

0.05y = 0.4 + 0.01y

y = 10 ml ---------- (4)

Adding (4) to (1),

The total amount of HCl required: 10 + 20 = 30 ml.



My questions is...what is the "simple proportion" mean...hows and where was it derivived from...

Thanks

 

[dlink]

First, you need to realize that 0.05 M HCl gives rise to 0.05 M H+ (i.e. 0.05 mol of H+ in 1000 ml of solution).

If 1000 ml of solution gives you 0.05 mol of H+, then, the number of mol of H+ in y ml of solution will lead to:

(y * 0.05)/1000 ---> the numerator

Now, before adding in the y ml of HCl, the neutralized solution is already known to be 20 ml of HCl plus 20 ml of KOH, which is 40 ml of solution. With y ml added to make it a final solution of pH 2, it will amount the solution volume to (40 + y) ml. But remember you'll need to express the volume of solution in liter. That leads to:

(40 + y)/1000 -----> denominator

By defination, the concentration of H+, or [H+] is expressed as the number of mole of H+ divided by volume of solution in 1 liter, or 1 dm3.

numerator/denominator = 1 x 10-2

Substituting the above relation leads to y = 10 ml, Adding it to the first added amount of 20 ml gives you the final answer of 30 ml.

Does the above help or is there anything else you're still confused with?

 

[paradigm2200]

ok...that makes sense....thanks

 

[joooj86]

Kaplan never emphasized or tested this concept....there is not a single question about this in kaplan or topscore...i decided not to worry about it..was that a bad decision or an acceptable one?

 

[dlink]

Kaplan never emphasized or tested this concept....there is not a single question about this in kaplan or topscore...i decided not to worry about it..was that a bad decision or an acceptable one?

Certainly not to worry about it in the real DAT, joooj86. But you should treat this question as means to further strengthen your molar/stoichiometry reasonings. 🙂