Question

-Calculate the pH of the solution obtained by mixing 25.00 mL of 0.1000 M NaOH with 25.00
mL of HClO 0.1000 M.

-Calculate the pH of the solution obtained by mixing 25.00 mL of 0.1000 M HCl with 25.00
mL of NaOCl 0.1000 M

Answer 1

In the first case we are adding the same volume and concentration of both, so we may safely assume that what we have in our final solution is NaClO (and some more water that is also formed). This is a salt that hydrolizes according to the equilibrium:

$ClO^{-}+H_{2}O\leftrightharpoons HClO+OH^{-}$

So it will result in a basic pH. In order to calculate the pH value, we have to use the following chart to follow the concentration of all species. Bear in mind that since the final volume of solution is double the volume of each of the mixed solutions, the concentration of the product is half the initial concentrations; that is: 0.0500M.

	ClO-	HClO	OH-
initial	0.0500M	0	0
change	-x	+x	+x
equilibrium	0.0500M-x	x	x

Now we can include this values in the expression for the basicity constant of ClO^-, which can be calculated from the Ka of HClO (2.9x10^-8), using:

$Ka\cdot Kb=Kw=10^{-14}$

So:

$Kb=\frac{10^{-14}}{Ka}=\frac{10^{-14}}{2.9x10^{-8}}=3.57x10^{-7}$

The expression for Kb is:

$Kb=3.57x10^{-7}=\frac{[HClO][OH^{-}]}{[ClO^{-}]}=\frac{x^{2}}{0.0500M-x}$

This is a quadratic equation that can be solved to yield:

$x_{1}=0.000133;x_{2}=-0.000133$

The first one is the only solution which makes chemical sense, so it is the one we will use. Since x was the concentration of OH^-, we can calculate:

$pOH=-log([OH^{-}])=-log(0.000133)=3.88$

And:

$pH=14-pOH=14-3.88=10.12$

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If we mix equal amounts of sodium hipochloride and HCl, we will obtain full reaction and get HClO and NaCl as products. HClO is a weak acid, which dissociates:

$HClO\leftrightharpoons ClO^{-}+H^{+}$

And we can calculate the pH in a manner similar to the one we used in the previous item. The initial concentration is again 0.0500M, and the chart to calculate the equilibrium concentrations would be:

	HClO	ClO-	H+
initial	0.0500M	0	0
change	-x	+x	+x
equilibrium	0.0500M-x	x	x

And now we use the expression of Ka for the calculation:

$Ka=2.9x10^{-8}=\frac{[H^{+}[ClO^{-}]}{[HClO]}=\frac{x^{2}}{0.0500M-x}$

The solutions for this quadratic equation are:

$x_{1}=0.0000381;x_{2}=-0.000381$

Again, the positive answer is the one we use, and we calculate pH as:

$pH=-log([H^{+}])=-log(0.0000381)=4.42$