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bugnsprout bugnsprout
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11 years ago
A buffer was made by mixing 0.1426 moles of HClO2 with 0.1565 moles of NaClO2- and diluting to exactly 1 liter. What will be the pH after addition of 10.00 mL of 0.2579 M HNO3 to 50.00 mL of the buffer? Ka(HClO2) = 1.100x10-2.  

I solved up the point where I get pH of 2.3177, but the answer is wrong. Am I missing a step or did I do my calculations wrong?
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#1
11 years ago
There are two steps to solving this. First ASSUME that a single direction reaction takes place between the strong acid and the ClO2- ion. Do some stoichiometry to figure out how much HClO2 is made. Then figure out what ClO2- is leftover after reacting with the strong acid. Be sure you recognize that you have 1/20 of the moles given in the 50 mL  because they were originally added to 1L not the 50mL.

After doing that stoich, then consider the Ka reaction for HClO2. You will need to use your answers from the stoich above as the initial values and then give a change of 'x' to find the equilibrium values (I call this filling out an ICE box with my students...) to put into the Ka expression. Be sure you account for the dilution (50mL + 10mL is the new volume).
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