Problem 15-123

HOCl H⁺ + OCl^-       K_a = 3.5x10^-8 = [H⁺][OCl^-]/[HOCl]

When the pH is 8:
3.5x10^-8 = [10^-8][OCl^-]/[HOCl]

Therefore to have a pH = 8 for this system, you need a ratio [OCl^-]/[HOCl] of 3.5
molar ratio or molar concentration is the same here, everything is in the same beaker.

To generate OCl^-: acid-base reaction: IRF table
(Table in mole, or for one Litre of solution)

	HOCl	+	NaOH		NaOCl	+	H2O
I	0.05		b		0		excess
R	-b		-b		'+b		'+b
F	0.05-b		0		b		excess

The strong base is all consumed creating the corresponcing number of mol of conjugate base OCl^-

[OCl^-]/[HOCl] = 3.5 = b/(0.050-b)

- Search for b and you will know how many mol of NaOH to add in one liter of solution.
- If you know the mol of NaOH, you know the volume of NaOH to add since V = n/molarity
- No need to know the exact concentration of chemical since it is a molar ratio problem.

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