These exercises deal with logarithmic scales. The hydrogen ion concentration of a sample of each substance is given. Calculate the pH of the substance.
\[Ph=-\log(H^+)\]?
if i remember but this may be wrong. if so you get for the first one \[-\log(5\times 10^{-3})=\log(5)-\log(10^{-3})=-\log(5)-(-3)=3-\log(5)\]
that is the full process when you show work for number 1?
yes. assuming the formula i wrote is correct. i get about 2.3 as a decimal
actually the answer is right but i made a typo. it should be \[-\log(5\times 10^{-3})=-\log(5)-\log(10^{-3})=-\log(5)-(-3)=3-\log(5)\]
but i repeat make sure the formula is right. you did not post it, i just looked it up.
you are correct in that formula.
what did you get for the other 2?
next one would be \[4-\log(3.2)\] and last one would be \[9-\log(5)\]
would it be to much to ask for how you got those. like could you show me the work
for those 2
cause it says that b) should be 3.5 and c) should be 8.3
yes that looks right. i just gave you the formula. if you want a decimal you have to use a calculator.
can you show me the steps for b and c
\[4-\log(3.2)=3.494...\] \[9-\log(5)=8.3...\]
sure it is the same as before
how did you get to that point for those 2 i mean i need to be able to write it all out
you have \[pH=-\log(a\times 10^b)=-(log(a)+\log(10^b)=-(\log(a)+b)=-b-\log(a)\]\] \[=
lets look at it simply
\[\log(5\times 10^{-9})=\log(5)+\log(10^{-9})=\log(5)-9\] are those steps clear?
first step because \[\log(ab)=\log(a)+\log(b)\] aways
second step because \[\log(10^x)=x\]
now you want the negative of that, and the negative of \[\log(5)-9\] is \[9-\log(5)\]
so if i want the pH of \[7\times 10^{-11}\] i go right to the answer of \[pH=11-\log(7)\]
of course i need a calculator to find the decimal
10.155
ok fine. i just made up that example but i hope you see how it works. except for the actual calculator work you can almost do it in your head
thanks for your help
i have one more that estudier cant figure out
thats my only other question i have no idea how to do it
where? i will look
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