Question 16.16: A solution made by adding solid sodium hypochlorite (NaClO) ...

Analyze NaClO is an ionic compound consisting of Na^{+}   and   ClO^{-} ions. As such, it is a strong electrolyte that completely dissociates in solution into Na^{+}, a spectator ion, and ClO^{-} ion, a weak base with K_{b} = 3.3 × 10^{-7} (Equation 16.37). Given this information, we must calculate the number of moles of NaClO needed to increase the pH of 2.00 L of water to 10.50.

Plan From the pH, we can determine the equilibrium concentration of OH^{-}. We can then construct a table of initial and equilibrium concentrations in which the initial concentration of ClO^{-} is our unknown. We can calculate [ClO^{-}] using the expression for K_{b}.

Solve
We can calculate [OH^{-}] by using either Equation 16.16 or Equation 16.20; we will use the latter method here:

K_{w} = [H_{3}O^{+}][OH^{-}] = [H^{+}][OH^{-}] = 1.0 × 10^{-14}      (at   25  °C)         [16.16]

pH + pOH = 14.00      (at  25  °C)                           [16.20]

pOH = 14.00 – pH = 14.00 – 10.50 = 3.50

This concentration is high enough that we can assume that Equation 16.37 is the only source of OH^{-}; that is, we can neglect any OH^{-} produced by the autoionization of H_{2}O. We now assume a value of x for the initial concentration of ClO^{-} and solve the equilibrium problem in the usual way.

We now use the expression for the basedissociation constant to solve for x:

x =\frac{(3.2 × 10^{-4})²}{3.3 × 10^{-7}}+ (3.2 × 10^{-4}) = 0.31 M

We say that the solution is 0.31 M in NaClO even though some of the ClO^{-} ions have reacted with water. Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water.