For 0.15 M HCN with Kc = 4.9 x 10^-10, what is the hydrogen ion concentration at equilibrium?

1. Let x be the concentration of each product formed, so the acid sits at 0.15 minus x. 2. The exact statement is 4.9 x 10^-10 = x squared divided by (0.15 minus x). 3. The constant is tiny, so x should be negligible and 0.15 minus x can be read as 0.15. 4. That gives x squared = 0.15 x 4.9 x 10^-10 = 7.4 x 10^-11. 5. Take the square root: x = 8.6 x 10^-6 mol per litre. 6. Check the assumption: 8.6 x 10^-6 divided by 0.15 is 0.006 percent, far below 5 percent. 7. The approximation is therefore valid, and the quadratic gives the same two figures. 8. Quoting 7.4 x 10^-11 is the trap of stopping at x squared and never taking the root. _Source: OpenStax Chemistry (CC BY 4.0), Ch 13 "Fundamental Equilibrium Concepts", section 13.4 Equilibrium Calculations_