I was given the following problem:
what is the normality of a solution containing 6 grams of H2SO4 in a 200ml solution? my teacher obtained a normality of 39normals while I took into account that I had two hydrogen free and obtained a result of 0.6 N, who's right?
2 answers
.6 is correct ... due to H⁺ activity
Did you make a typo and you meant to type in that the teacher obtained 0.30 N?
Technically, you can't say. Both could be right but I disagree with Scott as to why that is so. You must know the equation involved. For example, if you titrate one(1) H then
mL x N x milliequivalent weight = grams.
200 x N x 0.098 = 6g and N = 0.31 which rounds to 0.3 N to 1 sig fig. and the equation is
NaOH + H2SO4 ==> NaHSO4 + H2O
BUT
if the equation is
2NaOH + H2SO4 ==> Na2SO4 + 2H2O then
200 x N x 0.049 = 6 and N = 0.61 which rounds to 0.6 to 1 s.f.
The equivalent weight of H2SO4 is 98 when 1 H is used but it is 98/2 = 40 when both H atoms are used.
Technically, you can't say. Both could be right but I disagree with Scott as to why that is so. You must know the equation involved. For example, if you titrate one(1) H then
mL x N x milliequivalent weight = grams.
200 x N x 0.098 = 6g and N = 0.31 which rounds to 0.3 N to 1 sig fig. and the equation is
NaOH + H2SO4 ==> NaHSO4 + H2O
BUT
if the equation is
2NaOH + H2SO4 ==> Na2SO4 + 2H2O then
200 x N x 0.049 = 6 and N = 0.61 which rounds to 0.6 to 1 s.f.
The equivalent weight of H2SO4 is 98 when 1 H is used but it is 98/2 = 40 when both H atoms are used.
1 answer