acetic acid pKa=4.76, ka__________ M

Question

asevilla5 · Answer

http://www.webqc.org/phsolver.php this might help you not sure

kainui · Answer

$$\LARGE pK_a=-\log_{10}(K_a)$$

kkutie7 · Answer

$$10^{-pKa}$$

kkutie7 · Answer

someone gave you the equation as 
$$pKa=-log(K_{a})$$
solve for Ka 
$$-pKa=log(K_{a})$$
$$10^{-pKa}=K_{a}$$
you know pKa plug it in and solve

toxicsugar22 · Answer

is it 1.737800829 10^-5

kkutie7 · Answer

yes

kkutie7 · Answer

I'm not sure sorry

australopithecus · Answer

pKa = -Log(Ka)

australopithecus · Answer

toxicsugar22 if you dont trust her alegbra you are can always look up rules for manipulating logarithms

australopithecus · Answer

I will tell you though that it is correct but you should know its correct.

australopithecus · Answer

your algebra is wrong

australopithecus · Answer

Look at kkutie7s answer she does the algebra correctly, http://www.mathwords.com/l/logarithm_rules.htm

kkutie7 · Answer

you get 10^(-pKa) because you need to get rid of the log on the right hand side so you do 10^ to each side to just get Ka.
as for sig figs... I can't remember the rules right now, but I usually put it as 1.74*10^(-5)

toxicsugar22 · Answer

is anyone formalize with the Henderson hasselbach formula

australopithecus · Answer

Yes it is very easy

toxicsugar22 · Answer

cna you apply this to thate qation and find ka from that

australopithecus · Answer

https://www.khanacademy.org/science/chemistry/acids-and-bases/v/buffers-and-hendersen-hasselbalch

toxicsugar22 · Answer

ph=pka+log{a-/{HA}

toxicsugar22 · Answer

do i get the same anwer and how would i do that

toxicsugar22 · Answer

or i cant use this fromula for acetic acid pKa=4.76, ka__________ M

australopithecus · Answer

I gave you the formula 

pKa = -Log(Ka)
Ka = 10^(-pKa)

toxicsugar22 · Answer

1.737800829 10^-5

australopithecus · Answer

http://www.wolframalpha.com/input/?i=Ka+%3D+10%5E%28-4.76%29

australopithecus · Answer

that is your answer