An industrial chemist studying bleaching and sterilizing prepares a hypochlorite buffer using 0.350 M HClO and 0.350 M NaClO. (Ka for HClO = 2.9 × 10−8) Find the pH of 1.00 L of the solution after 0.030 mol of NaOH has been added.

Here we will first have to calculate the A⁻ formed in the 1. 0 L solution which is formed by the reaction of HClO with the strong base NaOH and add it to the original mol of NaClO

mol NaClO = mole NaCLO originally present in the 1L of M solution + 0.030 mol produced in the reaction of HCLO with NaOH

0.350 mol + .030 mol = 0.380 mol

New concentrations :

HClO = 0. 350 mol-0.030 mol = 0.320 M (have to sustract the 0.030 mol reacted with NaOH)

NaClO = 0.380 mol/ 1 L = 0.380 M

Now we have all the values required and we can plug them into the equation

pH = -log (2.9 x 10^-8) + log (0.380/.320) = 7.45

adefioyelaoye adefioyelaoye · Answer 2 · 2022-03-25T14:03:50+01:00

The pH of 1.00 L of the solution is 7.45.

What is pH?

This is defined as the power of hydrogen and it measures how acidic or basic a substance is.

Using Henderson-Hasselbalch equation:

pH = pKa + log ((A⁻)/(HA))

mol of NaClO = mole NaCLO initially present in the 1L of M solution + 0.030 mol produced in the reaction of HClO with NaOH

0.350 mol + .030 mol = 0.380 mol

We can then calculate the new concentrations below:

HClO = 0. 350 mol-0.030 mol = 0.320 M

NaClO = 0.380 mol/ 1 L = 0.380 M

Substitute the values into the equation

pH = -log (2.9 x 10⁻⁸) + log (0.380/.320)

= 7.45

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