Distilled water was added to 24.0dm³ of sulphuric acid to give a total volume of 250cm³. When 25.0cm³ of this

Question

solution is titrated, it required 22.80cm³ of 1.325g/dm³ of NaOH. Calculate the original concentration of sulphuric acid. The answer is 0.157mol/dm³. Can anyone show me how to go about doing it?

Answer 1

i think you mean 24 cm3, not dm3

find the moles of NaOH reacted first

mass of NaOH reacted = 22.80/1000 x 1.325 = 0.03021 g

Mr of NaOH = 23 + 16 + 1 = 40

moles of NaOH reacted = 0.03021/40 = 7.5525 x 10^-4 moles


We know 1 mole of NaOH would react with 1/2 mole of H2SO4

Moles of H2SO4 present in the 25 cm3 = (7.5525 x 10^-4)/2 = 3.77625 x 10^-4

Moles of H2SO4 present in 250cm3 = 250/25 x 3.77625 x 10^-3 = moles of H2SO4 present in 24cm3 of original solution

Moles of H2SO4 present in 1 cm3 of original solution = 3.77625 x 10^-3 / 24 = 1.5734375 x 10^-4 moles

1 dm3 = 1000 cm3

moles of h2so4 in 1 dm3 of the original sol = 1000 x 1.5734375 x 10^-4 = 0.157 mol/dm3

Answer 2

jus find the specigravity of concentrated sulphuric acid from any refrence table than find what amount in gram of sulphuric acid is there in 24.0dm cube than find normality of the solution and than as usual find normality of NaOH solution than use normality equation to find the answer.

Answer 3

i'll give u a hint...well it would be if u didn't already know it but
1dm^3 is 1 L and
1cm^3 is 0.001L

ohand that concentration formula
cv=n