quick chem q

[grami001]

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A solution that is formed by combining 200 ml of 0.15 M HCl with 300 ml of 0.090 M NaOH has an OH- concentration of
a. 1.0 x 10-7
b. 1.7 x 10-12
c. 0.030
d. 0.0030
e. 3.3 x 10-12

The answer is B but I do not know how to do it. Can anybody explain me how to solve this type of question? Thanks

 

[osimsDDS]

ive been staring at this question for 5 min and dont have a clue how to do it...im curious!!! someone post already

 

[baedero1]

HCl --> 0.2L*0.15M = 0.03mol
NaOH --> 0.3L*0.09M = 0.027mol

0.03-0.027 = 0.003mol of H+

0.003mol/0.5L = 0.006M of H+

pH = -log0.006M = 2.22

so, pOH = 14-2.22 = 11.78

[OH] = 10^-11.78 = 1.66*10^-12

 

[grami001]

thanks baedero

 

[osimsDDS]

I THINK I GOT IT AFTER TRYING TO FIGURE IT OUT FOR 15 MIN!!!!

ok so you have .15M HCl x .2L = .03 moles HCL and then you have .09M NaOH x .3L = .027 moles NaOH...now I think you subtract .03-.027=.003 moles but i dont know why you would do this thats where I am getting lost..

Then my logic tells me to divide .003 by .5L of the new combined solution...this gives you .006M of the new solution

Now you use Kw=[h30][oh-] to find oh- concentration...you get 1x10^-14/.006= 1.667 x10^-12

 

[osimsDDS]

Question: why do you subtract the moles, logically someone give me an answer...what is the reason behind subtracting .03-.027? thanks

 

[baedero1]

yes, i subtract to get final pH of that solution.
remember HCl and NaOH are strong acid and base. after neutralization reaction, H+ is remained in that solution. based on that idea, you can know the pH and also pOH of that solution.

 

[osimsDDS]

uhhh a strong acid and a strong base make a salt and water

HCl + NaOH ----> NaCl + H2O

Are you sure a strong base and a strong acid produce H+ cuz from this it doesnt seem like it....

 

[osimsDDS]

NVM I got it, its because .03 moles of HCl is going to react with .027 moles of NaOH (limiting reactant) and the left over HCl moles in solution will be .03-.027....after neutralization of the reaction you will have .003 moles of HCl still in solution so that is where you get H+ concentration...but instead of doing it your way to find the pOH and all that you can just use the kw=1x10^-14 equation...much faster and easier to find OH- concentration

 

[DDSguyLA]

this is that one question i'm skipping on the test.

 

[Tina324]

Hey guys..great question.

Instead of subtracting the moles of NaOH from the moles of HCl..why can't we subtract the Molarity of NaOH(.027 mol/.5 L=.054M) from the Molarity of HCl(.03 mol/.5L=.6M)? Can someone please explain theoretically why we can't use Molarity?

Thank u!!!

 

[Sagar Chauhan]

hey Tina,
Yes you can subtract the Molarity of NaOH from Molarity of HCL, but first of all Molarity of HCL would be 0.03mol/0.5L = 0.06M not 0.6M as you wrote in the question.
So, when we do the subtraction, we get 0.06-0.054=0.006M of HCL.
Now, we can use the formula
[H+]*[OH-]=Kw

0.006 * [OH-] = 1*10^(-14)
[OH-]= 1*10^(-14)/0.006
[OH-]= 1.67*10^(-12)

In this way, we don't need to do pH and pOH work and we could save our time!!

 

[Tina324]

thanks sagar!