what-would-be-the-equilibrium-concentrations-of-the-species-present-in-a-2-50-m-

Question

For a 2.50 M NaClO solution, what would be the concentration of the species present at equilibrium?

Accepted Answer

[$$Na^{+}$$] = 2.50 M; [$$ClO^{–}$$]  = 2.50 M; [HClO] = 9.28×$$10^{–4} M$$; [$$OH^{–}$$] = 9.28×$$10^{–4} M$$; [$$H_{3}$O^{+}$$] = 1.08×$$10^{–11} M$$

Answer

[$$Na^{+}$$] = 2.50 M; [$$ClO^{–}$$]  = 2.39 M; [HClO] = 2.98×$$10^{–4} M$$; [$$OH^{–}$$] = 2.98×$$10^{–4} M$$; [$$H_{3}$O^{+}$$] = 8.10×$$10^{–11} M$$

Answer

[$$Na^{+}$$] = 1.25 M; [$$ClO^{–}$$]  = 1.25 M; [HClO] = 6.57×$$10^{–4} M$$; [$$OH^{–}$$] = 6.57×$$10^{–4} M$$; [$$H_{3}$O^{+}$$] = 1.52×$$10^{–11} M$$

Answer

[$$Na^{+}$$] = 1.25 M; [$$ClO^{–}$$]  = 1.25 M; [HClO] = 6.57×$$10^{–4} M$$; [$$OH^{–}$$] = 1.52×$$10^{–11} M$$; [$$H_{3}$O^{+}$$] = 6.57×$$10^{–4} M$$