An industrial chemist studying bleaching and sterilizing prepares several hypochlorite buffers. Find the pH of the following buffers.

(a) 0.090 M HClO and 0.090 M NaClO




(b) 0.090 M HClO and 0.135 M NaClO




(c) 0.135 M HClO and 0.090 M NaClO




(d) One liter of the solution in part (a) after 0.0050 mol NaOH has been added.

I only need help with (d). I got 8.07, but that was incorrect

Use the Henderson-Hasselbalch equation.

For A.
pH = pKa + log (base)/(acid)
pH = pKa + log(ClO^-)/(HClO)
pH = pKa + log(0.09)/(0.09)
pH = pKa + log 1
pH = pKa.

Is the pKa you're using 7.53? If not be sure and correct the following for your value of pKa. If that's the right answer 7.53 should be pKa.

...........HClO + OH^- ==> ClO^- + H2O
I..........0.09....0........0.09
add..............0.005.............
C........-0.005.-0.005....+0.005
E..........0.085...0........0.095

pH = pKa + log (base)/(acid)
If I didn't make a math error (and pKa = 7.53), pH = 7.58

Yeah, I got (a). The answer was 7.53. But when I try to solve (d), I keep getting the wrong answer. I subtracted the moles of NaOH from the acid and added the moles to the base. Then I did Ka = (x*([NaClO]+x))/([HClO - x) and then I found the pH. I got 8.07, but it was wrong.

I have no idea why I was having such a hard time with that one. Thank you so much for your help.

To determine the pH of the solution after adding 0.0050 mol NaOH, we need to consider the reactions that occur.

NaOH is a strong base that will react with the weak acid, HClO, to form water and the conjugate base, ClO-. The chemical equation for this reaction is:

HClO + NaOH -> H2O + NaClO

We can use the balanced equation to determine the change in moles of HClO and NaClO. Since NaOH and HClO react in a 1:1 ratio, adding 0.0050 mol NaOH will cause 0.0050 mol of HClO to react.

To find the new concentration of HClO, subtract 0.0050 mol from the original 0.090 M concentration.
c(HClO) = 0.090 M - 0.0050 mol / 1 L
c(HClO) = (0.090 M - 0.0050 M)

Similarly, we can find the new concentration of NaClO. Since HClO and NaClO have a 1:1 ratio, the decrease in HClO concentration also corresponds to a decrease in NaClO concentration.

c(NaClO) = 0.090 M - 0.0050 M

To obtain the pH of the solution, we need to find the concentration of H+ ions. Since HClO is a weak acid, it partially dissociates in water to form H+ ions and ClO- ions. The dissociation constant, Ka, for HClO is 2.9 x 10^-8.

The equilibrium expression for the dissociation of HClO is:
HClO ⇌ H+ + ClO-

Using the known concentration of HClO, we can calculate the concentration of H+ ions using the equation for the dissociation of HClO and the Ka value.

[H+] = √(Ka * c(HClO))

Once we have the concentration of H+, we can calculate the pH using the equation:
pH = -log[H+]

Substitute the correct values into the equations, and you should be able to obtain the correct pH for the solution in part (d).