Enigmatic Eve
Member
Does anyone know for the multiple choice question, whether we had to find the conjugate acid or conjugate base og HPO4^-??
So the answer is H2PO4-?? I think I got that....sig said:pH of H2SO4 is C - 1.6
Diprotic therefore -log[2 x 0.0115] = 1.64
and the conjugate acid one was H2PO4- wasn't it?
hmm.... I still chose 1.9 knowing full well that it was diprotic. The reason being that the second ionisation stage is in equilibrium, so we cant assume complete ionisation like the first stage. Guess we have to wait.sig said:pH of H2SO4 is C - 1.6
Diprotic therefore -log[2 x 0.0115] = 1.64
Ooh.. very smart.... I'll use what you said to talk myself out that I got that question wrong.. LolHeinz said:hmm.... I still chose 1.9 knowing full well that it was diprotic. The reason being that the second ionisation stage is in equilibrium, so we cant assume complete ionisation like the first stage. Guess we have to wait.
I hated that question.. didn't understand what it was going on about... so I made up crap.. hopefully they'll give me at least one mark out of sympathy. Does anyone know the correct answer to the question?persephone said:you're scaring me...move on to another question....like the standard potential thing....
I believe, however, that the question stated that the scale had already been 'zeroed' with the alkaline solution on it.coca cola said:30g = 30 ml (since the Q specifies that 1 ml/g)
therefore the total volume in the flask is 30ml, but 20ml is already occupied by the NaOH solution already in there.
therefore there must be only 10ml of H2SO4 added.
volume ratio is 2:1, mole ratio is 1:2. therefore molarity must be equal.
exactly the same as youHeinz said:hmm.... I still chose 1.9 knowing full well that it was diprotic. The reason being that the second ionisation stage is in equilibrium, so we cant assume complete ionisation like the first stage. Guess we have to wait.