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# How would you use the Henderson-Hasselbalch equation to calculate the pH of a solution that is 0.120 M in HClO and 0.185 M in KClO?

##"pH" = 7.65##

The **Henderson - Hasselbalch equation** allows you to calculate the of that contains a weak acid and its conjugate base by using the concentrations of these two species and the ##pK_a## of the weak acid.

##color(blue)("pH" = pK_a + log( (["conjugate base"])/(["weak acid"])))##

In your case, the weak acid is hypochlorous acid, ##"HClO"##. Its conjugate base, the hypochlorite anion, ##"ClO"^(-)##, is delivered to the solution by one of its salts, potassium hypochlorite, ##"KClO"##.

The acid dissociation constant, ##K_a##, for hypochlorous acid is equal to ##3.5 * 10^(-8)##

Now, before doing any calculations, try to predict what you expect the solution's pH to be compared with the acid's ##pK_a##.

Notice that when you have **equal concentrations** of weak acid and conjugate base, the log term will be equal to **zero**, since

##log(1) = 0##

This tells you that if you have **more conjugate base** than weak acid, the log term will be greater than ##1##, which will cause the pH to be **higher** than the ##pK_a##.

With this in mind, plug in your values into the H-H equation to get

##"pH" = -log(3.5 * 10^(-8)) + log( (0.185color(red)(cancel(color(black)("M"))))/(0.120color(red)(cancel(color(black)("M")))))##

##"pH" = 7.46 + 0.188 = color(green)(7.65)##