Suppose you want to prepare a hypochlorous acid (Ka = 3.0x10-8) buffer by mixing hypochlorous acid and sodium hypochlorite. If you want to prepare 1.0 L of buffer with a pH of 7.0 and a total buffer concentration of 0.10 M, how many moles of hypochlorous acid should be added? moles how many moles of sodium hypochlorite should be added? moles

0.0768 moles of hypochlorous acid, 0.0232 moles of sodium hypochlorite.

Explanation:

It is possible to answer this question using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [A⁻] / [HA] (1)

Where A⁻ is conjugate base, in this case sodium hypochlorite and HA wek acid (hypochlorous acid).

pka is -log ka; -log 3.0x10⁻⁸ = 7.52 and pH is desire pH, 7.0

As you want to prepare 1.0L of a buffer 0.10M, you need 0.10mol of buffer, that means:

[A⁻] + [HA] = 0.10 (2)

Replacing (2) in (1):

7.0 = 7.52 + log₁₀ [0.10 - HA] / [HA]

0.302[HA] =  [0.10 - HA]

1.302[HA] = 0.10

[HA] = 0.0768 moles of hypochlorous acid

Replacing in (2):

[A⁻] + 0.0768 = 0.10

[A⁻] = 0.10 - 0.0768 = 0.0232 moles of sodium hypochlorite

I hope it helps!

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i believe the answer is 5,000