Chemistry - buffer problem

[tree2323]

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ok this is from my homework. Could someone walk me through how to figure this out - step by step...I don't want someone to just tell me the answer. I have 4 more of these to figure out on my own!

Here's the first:
A solution of sodium hypochlorite is made by dissolving the solid salt in 500mL of water. The pH of the resulting solution is found to be 4.5. The Ka of hypochlorous acid is 3 x 10^-8.

How much solid sodium hypochlorite (52.5 g/mol) must have been added to the water?



Thank you!!!!!!!!

 

[Rednawz]

umm ok, so i think the way you do this is as follows.

Ka=(H+)(ClO-)/(HClO). You know that H+=ClO- because for every molar of H you get 1 of HClO. You can figure out H+ from the pH using -log. So say the answer to that equation is x.

You know Ka, you know X, you want initial concentration, which is (initial that is unknown)-x, and oyu know x.

So 3x10^-8=x^2/Unknown-x
Again, you know X from the -log. Hope the use of variables isnt too confusing and tell me if it works out.

 

[tree2323]

ok, using the pH of 4.5, I figured out [H+] = 3.1*10^-5. (Right?)

So, if I'm just plugging in the Ka value and the 3.1*10^-5 for both H+ and ClO- I get..... [HClO] = 0.032M.

And to find the answer in grams..
0.032(mol/L) * 0.5 L * 52.5(g/mol) gives me 0.84 grams.

Am I doing this right?!? thanks for helping 🙂

 

[tree2323]

ahhhh so the second problem I have has 2 pKa values.. I'm not sure what I'm doing with them both

 

[BerkReviewTeach]

Here's the first:
A solution of sodium hypochlorite is made by dissolving the solid salt in 500mL of water. The pH of the resulting solution is found to be 4.5. The Ka of hypochlorous acid is 3 x 10^-8.

How much solid sodium hypochlorite (52.5 g/mol) must have been added to the water?

It looks like you are missing some part of the question here. As you've written the question, you've added just NaClO (a weak base) to 500 mL of water (neutral species) and created an acidic solution (pH of 4.5 is acidic). There had to be some HClO or some hydronium ion in the solution initially for this question to work.