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You want to have a ph of 9.55 with a buffer. How many grames of CH3NH3ClO4 would you add to 800 ml of 0.258 M CH3NH2 to do this. Assume volume remains constant. Kb for methylamine is 4.2x10^-4. My answer is 320g. Can anyone check this please? I have an exam on buffers very fast. You would do me a great favor. I used Henderson to find concentration of methylamine required. Then the number of moles and grams.
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