didnt u have 20.2 before or i saw it wrong? we set up our equations differently, i set the V as the additional volume needed and also the numbers are different. my fault.

didnt u have 20.2 before or i saw it wrong? we set up our equations differently, i set the V as the additional volume needed and also the numbers are different. my fault.

and see what u get... where v is the additional volume needed after the it reached pH 7. i know ur way might be simpler but i just did this by intuition.

and see what u get... where v is the additional volume needed after the it reached pH 7. i know ur way might be simpler but i just did this by intuition.

because i did this problem in 2 two steps. first step was to add 20 ml to get to ph=7 and then from there, i calculated how much hcl you need to get from ph=7 to ph=2. thats how i did it

because i did this problem in 2 two steps. first step was to add 20 ml to get to ph=7 and then from there, i calculated how much hcl you need to get from ph=7 to ph=2. thats how i did it

oh i forgot to multiply by 40ml(.040L) to 10^-7. but anyways, you have to add that because at that point the solution is neutral(pH=7). So there is 10^-7(.040) moles of H+ in the solution.

Thats great, im an UCLA undergrad and UCLA is my dream school. hopefully i can get in but we'll see.

true but you can't add it just because it won't affect it much.

(1V)/(V+40) = 10^-2

or

(1V)/((V+20) + 20) = 10^-2

means

(moles of H+ left) divided by the total volume = concentration of 10^-2 M

V+20 is the volume of H+
20 is the volume of OH-

UCLA is a great school. I went to a party school for my other degrees. I partied too much. I'll probably have a hard time adapting to the UCLA lifestyle

im not saying that... im saying 10^-7 and 10^-7x40ml arent much different since its so small. you have to add 10^-7x40ml term to be truly correct because according to ur equation there is no H+ moles in the solution at ph=7 which is not true. you have to consider the number H+ moles there are at ph=7.

Although i hear ucla dental school is tough, you should be able to handle it..

im not saying that... im saying 10^-7 and 10^-7x40ml arent much different since its so small. you have to add 10^-7x40ml term to be truly correct because according to ur equation there is no H+ moles in the solution at ph=7 which is not true. you have to consider the number H+ moles there are at ph=7.

why is solution pH =7? its because at ph =7, the concentration of H+ in the solution is 10^-7M. This means that 40 ml of solution will have (10^-7M*40ml) moles of H+. Maybe u dont understand what i did exactly,

#of moles of H+ at ph=7 in the solution-> (10^-7Mx40ml) + (1MxV)<-#of H+ that will be added ------------------------
V + 40 <- u know that

why is solution pH =7? its because at ph =7, the concentration of H+ in the solution is 10^-7M. This means that 40 ml of solution will have (10^-7M*40ml) moles of H+. Maybe u dont understand what i did exactly,

#of moles of H+ at ph=7 in the solution-> (10^-7Mx40ml) + (1MxV)<-#of H+ that will be added ------------------------
V + 40 <- u know that

Do we also have this many moles of OH- in the neutral solution: 10^-7 * 40 ml? (i.e. pOH = 7 too) If we do doesn't that cancel 10^-7 * 40 ml H+. I think we need to find the extra H+ right?

You're right that there is 10^-7x40 ml of OH- in the solution. but you cant say they cancel out because equal number are present. ph=7 means that there are (10^-7x40ml) are dissociated (and same as for OH- as a result). So we need to include it. That's what i think. Does the topscore have your explanation? cuz ihavent taken topscore yet.

You're right that there is 10^-7x40 ml of OH- in the solution. but you cant say they cancel out because equal number are present. ph=7 means that there are (10^-7x40ml) are dissociated (and same as for OH- as a result). So we need to include it. That's what i think. Does the topscore have your explanation? cuz ihavent taken topscore yet.

I don't have topscore. Now, I think you are absolutely right. We usually ignore 10^-7 because it's usually a small number (relatively). In this case (and most other cases), 10^-7 doesn't affect the final answer much but technically & scientifically you're right. Thanks for the explanation.

Here is an interesting problem:

We add 10^-8 moles of H+ to 1 L solution. What is the pH? (assuming volume stays at 1 L)

assuming the 1L solution is at ph=7, then the pH of the new solution would be -log (10^-7+10^-8)/1. if theres some trick to it, this would probably be wrong.

assuming the 1L solution is at ph=7, then the pH of the new solution would be -log (10^-7+10^-8)/1. if theres some trick to it, this would probably be wrong.

We can also deduce that since it only takes .4 mL of HCl to drop the pH to 2 after reaching the isoelectric point at pH 7, the principle of a strong base titrated with strong acid. You know should know that it doesn't take much to lower the pH drastically after the isoelectric point is reached.