A 0.552 g sample of ascorbic acid(Vitamin C) was dissolved in water to a total volume of 20.0 ml and titrated with 0.1103 M KOH. The equivalence point occurred at 28.42 ml. The pH of the solution at 10.0 ml of added base was 3.72. From this data, determine the molar mass and Ka for vitamin C.
All I got do far is..
Vitamin C+ H2O-->
I don't know what to do?!:0
2 answers
if something has a pH of 3.72 its not a base... did you mean pOH?
28.42mL of O.1103mol/L KOH titrated / reacted with all the Vit C there was.
O.1103mol/L * 28.42mL *1L/1000mL = 3.13*10^-3mol KOH = 3.13*10^-3mol Vit C
O.552g/3.13*10^-3mol = ~176g/mol
To find Ka use pH=pka + log[base]/[acid]
pH is 3.72,
Initial [acid]=0.552g/176g/mol = 3.13*10^-3mol
Change is the amount of base added that will react with the initial amount of acid, = 3.13*10^-3mol - (0.1103M*.01L=1.103*10^-3mol) = 2.03*10^-3mol
pH=pka + log[base][acid]
3.72 = pka + log(1.103810^-3(/(2.03*10^-3)
pka = 3.72-(-0.2649...) = 3.984..
pka = -log[ka]
10^-pka = ka
10^-3.984... =ka = 1.04*10^-4
O.1103mol/L * 28.42mL *1L/1000mL = 3.13*10^-3mol KOH = 3.13*10^-3mol Vit C
O.552g/3.13*10^-3mol = ~176g/mol
To find Ka use pH=pka + log[base]/[acid]
pH is 3.72,
Initial [acid]=0.552g/176g/mol = 3.13*10^-3mol
Change is the amount of base added that will react with the initial amount of acid, = 3.13*10^-3mol - (0.1103M*.01L=1.103*10^-3mol) = 2.03*10^-3mol
pH=pka + log[base][acid]
3.72 = pka + log(1.103810^-3(/(2.03*10^-3)
pka = 3.72-(-0.2649...) = 3.984..
pka = -log[ka]
10^-pka = ka
10^-3.984... =ka = 1.04*10^-4