You ought to take a shot at it. First, consider how many protons (H+) you added when you dissolved the tablet in 50 ml of 0.5 M HCl. But you over-titrated the OH- by some amount because the solution was still acidic - if you had added exactly the same number of protons as you had hydroxide ions to begin with, the pH of the solution would have been 7. So now you add back enough OH- in 30.9 ml of 0.255 M sodium hydroxide to bring the solution to neutrality. That tells you how many more protons there were in 50 ml of .5M HCl than there were hydroxide ions in the original solution. Subtracting the extras from the total number of protons added tells you how many hydroxide ions there were, and since all of those ions must have come from the tablet, that will be the answer to the first part.
The second part might have more than one correct answer, but the simplest one is to subtract out the weight of the 5% binder from the 500 mg tablet leaving you with the weight of active ingredients. Then note that 2/5 of all the hydroxide ions must have come from Mg(OH)2 while 3/5 must have come from Al(OH)3.
Thank you for explaining this question so thoroughly !
I figured out how many moles there were in the tablet by subtracting nHCl by nNaOH, but I don't know how to get the moles of OH- from there
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