Question

A buffer solution was prepared by mixing `"392 mL"` of `"0.301 M"` `"NaClO"` and `"181 mL"` of `"0.281 M"` `"HClO"`. Calculate the `"pH"` of the solution?

Given that `K_a = 3.2 * 10^(-8)`. Express the answer rounded to 3 decimal places.

Answer 1

`"pH" = 7.860`

Explanation:

You're dealing with a weak acid - conjugate base buffer that has hypochlorous acid, `"HClO"`, as the weak acid and the hypochlorite anion, `"ClO"^(-)`, as its conjugate base.

As you know, the `"pH"` of a weak acid - conjugate base buffer can be calculated using the Henderson - Hasselbalch equation.

`"pH" = "p"K_a + log ((["conjugate base"])/(["weak acid"]))`

Here

`"p"K_a = - log(K_a)`

Now, your goal here is to figure out the concentrations of the hypochlorous acid and of the hypochlorite anion after you mix the two solutions.

Right from the start, you know that the volume of the resulting solution will be

`"392 mL + 181 mL = 573 mL"`

Now, use the molarity and the volume of the hypochlorous solution to calculate how many moles of the weak acid are present.

`181 color(red)(cancel(color(black)("mL solution"))) * "0.281 moles HClO"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.050861 moles HClO"`

Next, do the same for the sodium hypochlorite solution, which as you know, dissociates in a `1:1` mole ratio to produce hypochlorite anions.

`392 color(red)(cancel(color(black)("mL solution"))) * "0.301 moles ClO"^(-)/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.11799 moles ClO"^(-)`

To calculate the concentrations of the two species after the two solutions are mixed, use the total volume of the solution--do not forget to convert it to liters!

`["HClO"] = "0.050861 moles"/(573 * 10^(-3) quad "L") = "0.08876 M"`

`["ClO"^(-)] = "0.11799 moles"/(573 * 10^(-3) quad "L") = "0.2059 M"`

Plug your values into the Henderson - Hasselbalch equation to find the `"pH"` of the solution.

`"pH" = - log(3.2 * 10^(-8)) + log ( (0.2059 color(red)(cancel(color(black)("M"))))/(0.08876 color(red)(cancel(color(black)("M")))))`

`color(darkgreen)(ul(color(black)("pH" = 7.860)))`

The answer is rounded to three decimal places, the number of sig figs you have for your values.

Notice that the `"pH"` of the solution is higher than the `"p"K_a` of the weak acid, which happens because the buffer contains a higher concentration of the conjugate base than of the weak acid.

When this happens, the `"pH"` of the buffer will be higher than the `"p"K_a` of the weak acid. Similarly, when a buffer contains a higher concentration of the weak acid than of the conjugate base, then the `"pH"` of the buffer will be lower than the `"p"K_a` of the weak acid.