is the hydrogen ion concentration of a pH 3.8 solution higher or lower than that of a solution with a pH of 6.2?
wait explaing that a bit more if you dont mind
pH = - log(based 10) [H+] [H+] = 10^(-pH)
Use the second to find your hydrogen ion concentrations, you'll observe that the one with the lower pH has more hydrogen ions.
For the solution with a pH of 3.8, [H+] = 10^-3.8 = 1.58 x 10^-4 M For the solution with a pH of 6.2, [H+] = 10^-6.2 = 6.31 x 10^-7 M The solution with a pH of 3.8 has a higher hydrogen ion concentration.
how about if one solution has 100 times as many hydrogen ions as another solution, what is the difference, in pH units between the two solutions?
10 to the what power gives you 100? :P
2 :P
So heres another question if you wanna try it: If the pH of a solution is 8.5, what is the pH of a solution that has a 10000 times greater hydrogen concentration?
ok you got me but is the second question i asked you would that be 2 or negative 2?
It would be 2.
how so?
or how would i show work sorry if im being annoying i jus dont understand this topic
Its not your fault, I'm not really showing any work/explaining, sorry. Lets assume that solution A has a [H+] concentration greater than solution B. \[[H _{}^{+}]_{A}^{} = 10^{-pH _{A}^{}}\] \[[H _{}^{+}]_{B}^{} = 10^{-pH _{B}^{}}\]
If we divide the two equations, we get that \[[H _{}^{+}]_{A}^{} / [H _{}^{+}]_{B}^{} =10^{-pH _{A}^{} + pH _{B}^{}} = 100\]
ok i can see that
\[10^{-pH _{A}^{} + pH _{B}^{}} = 100\] Take the log based 10 of both sides:\[\log_{10}( 10^{-pH _{A}^{} + pH _{B}^{}} = 100)\] \[{-pH _{A}^{} + pH _{B}^{}} = 2\]
oh i see it now thanks! i am getting it a little now but if you dont mind i have about two other problems that need to be completed, but i dont exactly want you to do it for me but help me lead myself to an answer if you dont mind?
Sure, I'll help :)
alright thanks!!!!
if solution a contains 1 x 10^-6 MH+ ions and solution B contains 1 x 10^-8 MH+ ions, which solution contains more H+ ions
This question is just about your understanding of negative exponents I guess. Convert the numbers from scientific notation to standard notation to see which one is bigger.
ok dang i feel dumb now lol but do you know the relationship between hydrogen ion contrecations to the acidity and alklinity (basicity) of solution?
An easier way to do this problem would be to look at the pH's of each solution. If you look at the [H+] equation and compare it to the values they give, you'll notice that the first solution has a pH of 6 whilst the second has a pH of 8.
so would the ph of 6 be greater right?
Higher hydrogen ion concentration = lower pH = more acidity. Lower hydrogen ion concentration = higher pH = more basicity.
Yup, the pH of 6 has more...
awesome!!!!!!!!!!!
You'll get the hang of these problems after a while. Remember the key concepts and those two equations.
HA is an acid that ionizes 10% in a solution. what is the H+ of a 0.01 M solution of HA? what is its pH?
would the ph be 1 or 2?
oh and thanks again for all the help
If 10% of the 0.01 M is ionized, how much H+ are there?
um yeah now ya got me
-2?
10% of 0.01 M is just 0.001 M, so that is you H+ concentration. pH = - log [H+] = - log (0.001) = 3
pH is often positive by the way, unless your dealing with a strong acid, or a super concentrated solution.
so would 3 be the pH or the HA? or the amount of H+?
pH = 3, H+ = 0.001 M
HA is just some arbitrary acid, H is for the hydrogen an acid has, and A is for acid.
when did you lean all this stuff?
or where?!
Last year, took ap chem :)
like a boss!!!!
alright so in this next problem it seems like its working backwards what is the H+ of a solution whose pH is 8? what is the OH-?
Use these equations to help you: pH + pOH = 14 pH = - log [H+] pOH = - log [OH-] [H+] = 10^-pH [OH- ] = 10^-pOH
So for the first part, [H+] = 10^-8 M. For the second part, you get pOH = 6, so then [OH-] = 10^-6 M
I've got an ap bio final tomorrow, so I gotta go to sleep. Try to solve your problems using the equations I gave you, good luck! :D
alright good luck on your final and thanks!
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