2 Gchem Q

[prsndwg]

Advertisement - Members don't see this ad

Hi guys..
Can someone help me out with these two..

I thought this one was C, but its A.. is there like a spectator ion game that I am not aware of?

I could guess and narrow it down to C/D.. but clueless of how to solve it..

thanks in advance

 

[alanan84]

For the first one, I got A. Al goes from 0 to +3 (I believe) meaning it gets oxidized and is the reducing agent.

For the second one, copy and paste that question into google and you should find a thread from SDN a few years back that answers it.

 

[dentrilla]

Hi guys..
Can someone help me out with these two..

I thought this one was C, but its A.. is there like a spectator ion game that I am not aware of?

I could guess and narrow it down to C/D.. but clueless of how to solve it..

thanks in advance

Hey,

Reducing agents get oxidized! So, start off by assigning oxidation states (if you dont know how to do this watch chad's video on electrochem). You'll realize that Al goes from Al(s) --> Al+3.. it has LOST electrons and therefore got oxidized which means its the reducing agent (whoever invented that terminology is an idiot😀) Al(s) is the atom which is losing electrons in this reaction, not its partner on the other side.

My guess for the 2nd question is:

Basically we gotta start this problem by working backwards.

(mol HCl left over, after neutralization) / ( total volume) = [H+] = 0.01

[(1MHCl * a LHCl) - (0.02mol OH-)] / (a LHCl + 0.02 LNaOH) = 0.01

a LHCl = 0.0204L = 20.4 mL

my guess is it's either B or E

 

[prsndwg]

Ok i got it..the 1st one is cleared... i still dont get the second oe though!😕

 

[toothdoc0121]

i actually just read a thread on the second one but i forgot the link so let me try and explain it...

HCL= .02L*1M= .02mol
NaOH= x L*1M= x mol
pH(2)= .01M

(x mol-.02 mol)/(x L + .02L)= .01M

so your answer would be 20.2mL

hope this helps

 

[dentrilla]

i actually just read a thread on the second one but i forgot the link so let me try and explain it...

HCL= .02L*1M= .02mol
NaOH= x L*1M= x mol
pH(2)= .01M

(x mol-.02 mol)/(x L + .02L)= .01M

so your answer would be 20.2mL

hope this helps

That equals 20.4, and the equation he used was the exactly the one that i used. You have to look at what you would normally do in a strong acid/strong base reaction. Both compounds dissociate completely, and the H+ and OH- react completely --> H20. When the limiting reagent is finished, the left over mols H+ or OH- ( in this case we know its H+ since the final pH=2) divided by your total concentration = [H+] which is 0.01 in this case (since pH=2) Look at my answer again and try to understand it by answering the following question.

Whats the pH of a solution of 20.4mL of 1M HCl and 20mL of 1M NaOH. The answer will obviously turn out to 2?

If you can answer that then you'll see how I derived my equation.

 

[yumi43]

aha! i had the exact same question about the 2nd problem. i posted a thread here yesterday and someone answered it, but i am still a little confused. here is the link: http://forums.studentdoctor.net/showthread.php?t=657556

for the first problem, i don't think you can call a product an ox/red agent. those should always come from the reactants side... and then ditto what everyone else said...

 

[prsndwg]

That equals 20.4, and the equation he used was the exactly the one that i used. You have to look at what you would normally do in a strong acid/strong base reaction. Both compounds dissociate completely, and the H+ and OH- react completely --> H20. When the limiting reagent is finished, the left over mols H+ or OH- ( in this case we know its H+ since the final pH=2) divided by your total concentration = [H+] which is 0.01 in this case (since pH=2) Look at my answer again and try to understand it by answering the following question.

Whats the pH of a solution of 20.4mL of 1M HCl and 20mL of 1M NaOH. The answer will obviously turn out to 2?

If you can answer that then you'll see how I derived my equation.

ok.. I figured it out.. Thanks

 

[dentrilla]

ok.. I figured it out.. Thanks

anytime👍