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hey...just a quick q on dilutions vs titrations
i know you use mv=mv for diluitions for ex 100ml 5M naoh is diluted to 200ml wat is final M-
you just use the mv= mv equation and get 2.5 M
but for titrations the kaplan book uses vn=vn n is for normality
and it says to use this equation to find the end point........for example
100 ml 1M naoh is titrated w/ 5M h3po4 to find the volume to reach the end point you use nv=nv (100)(1)=(15)(x) x = 6.7ml....now does this amount of h3po4 that gets us to the endpoint give us a pH of 7?
what im confused about is how far off is the endpoint and equivalence point and on the dat what are they going to ask for??? how do i find out what volume of the acid will equalize to a ph of 7??
i know you use mv=mv for diluitions for ex 100ml 5M naoh is diluted to 200ml wat is final M-
you just use the mv= mv equation and get 2.5 M
but for titrations the kaplan book uses vn=vn n is for normality
and it says to use this equation to find the end point........for example
100 ml 1M naoh is titrated w/ 5M h3po4 to find the volume to reach the end point you use nv=nv (100)(1)=(15)(x) x = 6.7ml....now does this amount of h3po4 that gets us to the endpoint give us a pH of 7?
what im confused about is how far off is the endpoint and equivalence point and on the dat what are they going to ask for??? how do i find out what volume of the acid will equalize to a ph of 7??
