Ka for HCN = 6.0×10^-10
Ka = [H+][CN-]/[HCN]
after X moles/l have dissociated the concentrations are-
[H+] = X
[CN-] = X
[HCN] = 0.70 – X
Since Ka is very small, we know X << 0.70, so we can assume [HCN] = 0.70 (to 2 significant figures)
6.0×10^-10 = X^2 / 0.70
4.2×10^-10 = X^2
X = 2.0×10^-5
Perhaps your textbook gives a different value of Ka.
1.9×10^-5 would be the correct answer if the value of Ka was 5.2×10^-10
1.9×10^-5 would be the correct answer if Ka = 6.0×10^-10 and the concentration of HCN was 0.60 M (not 0.70 M)
CHEMIStry*?
You need Ka to solve it. My book (Zuhmdahl, 5th ed.) says it is 6.2 X 10^-10.
HCN = H+ + CN-
Let "X" equal [H+] and [CN-].
Ka = [H+][CN-] / [HCN] = 6.2 X 10^-10
X^2 / 0.70 = 6.2 X10^-10
X^2 = 4.34 X 10^-10
X = ( 4.34 X 10^-10)^0.5
X = 2.08 X 10^-5
So [H+] = [H3O+] = 2.08 X 10^-5 M
The slight difference in answers is probably because your textbook has a different Ka value for HCN.
Ka for HCN = 6.0×10^-10
Ka = [H+][CN-]/[HCN]
after X moles/l have dissociated the concentrations are-
[H+] = X
[CN-] = X
[HCN] = 0.70 – X
Since Ka is very small, we know X << 0.70, so we can assume [HCN] = 0.70 (to 2 significant figures)
6.0×10^-10 = X^2 / 0.70
4.2×10^-10 = X^2
X = 2.0×10^-5
Perhaps your textbook gives a different value of Ka.
1.9×10^-5 would be the correct answer if the value of Ka was 5.2×10^-10
1.9×10^-5 would be the correct answer if Ka = 6.0×10^-10 and the concentration of HCN was 0.60 M (not 0.70 M)