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Chemistry 15 Online
OpenStudy (toxicsugar22):

WHAT ARE THE CONCENTRATIONS OF H30+ AND OH- in tomatoes that have pH oh 4.14

OpenStudy (toxicsugar22):

DO H30+ FIRST THEN OH-

OpenStudy (toxicsugar22):

so walk me throught the steps

OpenStudy (anonymous):

PH = 4.14 it is a acid solution

OpenStudy (anonymous):

Ph = 4.14< 7

OpenStudy (toxicsugar22):

so what is H30+

OpenStudy (toxicsugar22):

ANSWER

OpenStudy (toxicsugar22):

is is -.617

OpenStudy (toxicsugar22):

is it -log(4.14) to get H30+ I GOT -.617

OpenStudy (anonymous):

nnnnnnnoooooo PH = -log H3O+ and POH = -Log OH- H3O+ = e -PH and OH- = e-POH we have ph = 4.14 it is an acid selution so : H3O+ = e-4.14 POH = 14 - PH = 14-4.14=9.86 so OH- = e-9.86 since we have an acide solution so [H3O+] must be >>>>>>> [OH-] so we get that because e-4.14 >>> e-9.86 us i know ....... thanks

OpenStudy (jfraser):

to find pH from concentration, you take the -log(H3O+) to find concentration from pH, you have to do the inverse function, use \(10^{-pH}\)

OpenStudy (toxicsugar22):

so it is [H30+]=1*10^4.14 and [0H-]=1*10^9.86

OpenStudy (toxicsugar22):

IT IS RIGHT

OpenStudy (joannablackwelder):

Close. It is 10^-4.14 for [H3O=]

OpenStudy (toxicsugar22):

so fore [H30+]=1*10^-4.14 and what about [OH-]=

OpenStudy (joannablackwelder):

You will need to calculate the value for [H3O+] first. Can you put that in your calculator?

OpenStudy (toxicsugar22):

is it log ( then 10*-4.14)

OpenStudy (joannablackwelder):

No, the log cancelled out.|dw:1409770755201:dw|

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