A 36.0 mL sample of aqueous sulfuric acid was titrated with 0.250 M NaOH until neutralized. The residue was dried and then weighed with a mass of 861 mg. What is the molarity of the sulfuric acid solution?
Answer is 0.168M ...please and thank you 🙂
H2SO4 + NaOH -> Na2SO4 + 2H2O
Mole of Na2SO4 = 0.861/142 = 0.006 mole
The mole of H2SO4 required for complete neutralization is also 0.006 mole (1:1 ratio)
Molarity of H2SO4 = 0.006/0.036 = 0.168 M
1 more... sighhhh
When 37 g of Ca(OH)2 is added to 1000mL of 0.55M H2SO4, the resulting pOH is?
Answer is 13...thank you again for any help guys
Ca(OH)2 + H2SO4 -> CaSO4 + 2H2O
Mole of Ca(OH2) = 37/74 = 0.5 mole
Mole of H2SO4 = 0.55/1 = 0.55 mole
Thus H2SO4 is excess; mole of H2SO4 excess = 0.55 - 0.5 = 0.05 mole
Lets assume the dissociation of H2O and the 2nd dissociation of H2SO4 are negligible and of course there is no change in volume upon adding Ca(OH)2 (s)
[H+] = 0.05/1 = 0.05 M
pH = -log[H+] = -log(0.05) = 1.3
pOH = 14 - 1.3 = 12.7
Pretty close to the answer. I'm not sure if they made some assumptions for the answer though.
Let me know if you have any questions.
Best
