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Fizban140
Registered User, __BANNED USERS

Edit, need help again.

I can't figure these out, its multiple choice so I know I am not getting the correct answer.

4)20 mL of 3.00 M solution with 30 mL of a 5.40 M solutionb you get ?

I couldnt get an answer for this, you find the moles add that then divide by total volume? I am not sure. Either way my anwer wasn't listed.

5) 3.00 L of 3.01 M solution NaCl called solution A. 2.00 L of 2.00 M AgNO3 called solution B., mix together make solution C. Calculate conecntration in M of Na+ ions in C.

I tried this one a lot and couldnt get an answer. Its 1:1 I think so I just need to find the combined molarity but I wasn't too sure how to do that, no examples in the book.

6) Solution 1 100 ML of 3 M solution

Solution 2 50 mL 3 M

solution 3 10 mL of 3 M

All taken from 500 mL flask containing 3 M all of this is sodium carbonate

What colume in mL of water must evaporate from solution 1 in order to have concentration of 4.62 M?

7) Which of followin is not a conjugate acid base pair?

I really hate these.

A) H2SO4, SO4 2-

b) HNO3, NO3 -

C) HC2H3O2, CH3O2-

D) H2PO4 -, HPO4 2-

E) Hbr, Br-

I knowi ts going to be a change of H- which to me looks like E, D, C, B and A but maybe not A because it is 2H?

8) Calculate the pH of .044 M HCl solution

fff I did this one so many times and didn't get the right answer, none of these is an option so it could be that I guess.

I can't figure these out, its multiple choice so I know I am not getting the correct answer.

4)20 mL of 3.00 M solution with 30 mL of a 5.40 M solutionb you get ?

I couldnt get an answer for this, you find the moles add that then divide by total volume? I am not sure. Either way my anwer wasn't listed.

5) 3.00 L of 3.01 M solution NaCl called solution A. 2.00 L of 2.00 M AgNO3 called solution B., mix together make solution C. Calculate conecntration in M of Na+ ions in C.

I tried this one a lot and couldnt get an answer. Its 1:1 I think so I just need to find the combined molarity but I wasn't too sure how to do that, no examples in the book.

6) Solution 1 100 ML of 3 M solution

Solution 2 50 mL 3 M

solution 3 10 mL of 3 M

All taken from 500 mL flask containing 3 M all of this is sodium carbonate

What colume in mL of water must evaporate from solution 1 in order to have concentration of 4.62 M?

7) Which of followin is not a conjugate acid base pair?

I really hate these.

A) H2SO4, SO4 2-

b) HNO3, NO3 -

C) HC2H3O2, CH3O2-

D) H2PO4 -, HPO4 2-

E) Hbr, Br-

I knowi ts going to be a change of H- which to me looks like E, D, C, B and A but maybe not A because it is 2H?

8) Calculate the pH of .044 M HCl solution

fff I did this one so many times and didn't get the right answer, none of these is an option so it could be that I guess.

## Posts

Yes, the other problems used mL so I made sure, this is as written in the book is 12.5 g KNO3; 1.15 L

Your equation isn't balanced. It should be N2 + 3H2 --> 2NH3

If the volumes being mixed were equal, you could just calculate the average concentration from the two solutions. In this case you need to account for the fact that there's more of one solution than the other. 40% of your final 50 ml solution is 3 M, the other 60% is 5.40 M, so: (0.4 * 3M) + (0.6 * 5.40M) = your answer.

Keep in mind that molarity is a measure of concentration. You'd do the exact same calculation if you were mixing, say, solutions of 30% and 54% ethanol. Obviously the final concentration has to be somewhere between the concentrations of the original solutions; if it isn't, you'll know you've got it wrong.

That's the simplest way to calculate these. There are more complicated (and possibly more "correct" from a teacher's standpoint) ways to calculate this. For example, remembering that molarity (M) stands for moles/litre, you can do as you suggested yourself: find the total amount of moles, divide by final volume. In this case: (0.020 l * 3.0 mol/l + 0.030 l * 5.40 mol/l) / (0.020 l + 0.030 l). Just keep an eye on the units.

Note that they only ask for the concentration of Na+ ions. Unless I'm missing something, the silver nitrate is there just to confuse you. There is no Na in AgNO3 so solution B might just as well be water for the purposes of this exercise. You're simply diluting 3 liters of NaCl solution into a final volume of 5 l.

When diluting and concentrating solutions, remember that the total amount of moles in the solution stay the same, only concentration (c) and volume (v) change. So you start with the formula c1v1 = c2v2 and calculate the unknown term. In this case you need to find out final volume: v2 = (c1v1)/c2 = (3M * 0.1l)/4.62M. This gives you the volume of the 4.62 M solution. Then just subtract that from the original volume to see how much needs to evaporate.

You're right with your doubts about A. Sulfuric acid dissociates in two steps (H2SO4 -> HSO4- -> SO4 2-). The conjugate base of H2SO4 is HSO4-.

Take another look at C and see if you missed something there. =)

Just remember that HCl is a strong acid: all of it dissociates in solution. You ignore the H+ concentration of water, because it's many orders of magnitude below what we're looking at here. So the concentration of H+ is considered the same as the concentration of HCl. Remember that pH is just the negative 10-base logarithm of the H+ concentration, so you can calculate this simply as -log(0.044).