10+ Year Member
7+ Year Member
- Jul 9, 2008
Find the Ka of HCN, given that a 0.20 M solution of HCN([FONT=TimesNewRoman,Italic][FONT=TimesNewRoman,Italic]aq..) is 0.002 % ionized at 25 °C.
[H+][CN-] = Ka and .002%= 2.0x10^-5
(2.0x10^-5)2 =Ka doesn't seem to work. why is [H+] not equal to the given ionization ???
answer key is 8.0x10^-11, my answer was 2.0x10^-9. answer key always says [H+]=(2.0x10^-5)(2.0) instead of the given ionization...
thanks for the help..