Question

Calculate the PH of 0.175 M HY, if Ka=1.50x10^(-4)

Answer 1

so write an ionization expression: \(Ka=\dfrac{[H^+][X^-]}{[HX]}\)

make an ICE table, plug Ka and the values from E into the expression, solve for x.

in this example x=\([H^+]\), use \(pH=-log[H^+]\)

Answer 2

hi @aaronq, thank you. I realized i need to use the quadratic formula but for some reason my x values aren't giving me the right answer :|

Answer 3

did you get \([H^+]\)=0.005 M?

Answer 4

@aaronq somehow i got something totally different :/

Answer 5

how did you get your answer?

Answer 6

was this what you had? \(1.50*10^{-4}=\dfrac{x^2}{(0.175 -x)}\)

Answer 7

yes ! @aaronq

Answer 8

hm it's likely you made mistakes solving it, try again? if you dont get the right values i'll show you in steps.

Answer 9

hm :/ I used the quadratic formula, is that what you had used as well?

Answer 10

@aaronq

Answer 11

yeah, the set up was: \(x=\dfrac{\sqrt{42009}-3}{40000}\)

Answer 12

i had it set up this way...:\[\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]
\[\frac{ -(1.50\times10^{-4}) \pm \sqrt{(1.50\times10^{-4}) ^{2}-4(1)(2.626\times10^{-5})} }{ 2 }\]

Answer 13

i did it with wolfram, hold on i'll double check it

Answer 14

ok! this is where i was most stuck :|

Answer 15

ohhh you missed a negative sign, it's \(-2.626*10^{-5}\)

Answer 16

did i!? thats weird b/c i did try it with the negative as well and it didn't work for some reason. I think it may just be my work :/

Answer 17

should work !

Answer 18

it's easy to screw up the quadratic equation, dont feel bad lol

Answer 19

Ok @aaronq ill give it another try but i have the right approach correct?

Answer 20

Yeah, the exact equation is

\(x=\dfrac{(-1.50x10^{-4})+\sqrt{(1.50x10^{-4})^2-4*(-2.625*10^{-5}})}{2}\)

Answer 21

thank you so much! @aaronq

Answer 22

no problem, glad you got that sorted out!