Equilibrium Prob

[allstardentist]

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Find the Ka of HCN, given that a 0.20 M solution of
HCN(aq) is 0.002 % ionized at 25 °C.
A. 8 x 10–11
B. 4 x 10–8
C. 4 x 10–11
D. 8 x 10–8
E. 8 x 10–7

why is it A? i get 8x 10-9

 

[NilamPatel]

ka = [h3o+][conjugate base] \ [acid]

in this problem:

[h3o+] = .002% of .2
[conjugate base] = also .002% of .2
[acid] = .2

.002% of .2 = 4X10-6 -----> square it because it the concentration of both h3o+ and conjugate base and you get 16X10-12 and divide that 2X10-1 or .2 and you get -----> 8X10-11

 

[allstardentist]

yeah, ure right. i made a calculation error in converting percents to fraction.