The questions are solving for: You need to determine the concentration of a sulf
ID: 818310 • Letter: T
Question
The questions are solving for:
You need to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide solution. You have a 0.1379 M standard solution. You take a 25.00 mL sample of the original acid solution and dilute it to 250.0 mL. You then take a 10.00 mL sample of the dilute acid solution and titrate it with the standard solution. You need 13.88 mL of the standard solution to reach the endpoint. What is the concentration of the original acid solution? Concentration = A 60.00 mL solution of 0.4500 M AgN03. was added to a solution of AsO43-. Ag3AsO4 precipitated and was filtered off. Fe3+ was added and the solution was titrated with 0.2400 M KSCN. After 36.90 mL of KSCN solution had been added, the solution turned red. What mass of ASO43- was in the original solution?Explanation / Answer
Determination of H2SO4 solution concentration
H2SO4 + 2NaOH --> Na2SO4 + 2H2O
moles of NaOH used for 10 ml aliquot solution titration = molarity x volume
= 0.1379 M x 13.88 ml
= 1.914052 mmol
So,
moles of H2SO4 reacted = 1.914052/2 = 0.957026 mmol
molarity of H2SO4 in 10 ml aliquot = 0.957026 mmol/10 ml = 0.096 M
molarity of original acid solution = 0.096 M x 250 ml/25 ml = 0.96 M
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ASO4^3- + 3AgNO3 --> Ag3AsO4 + 3NO3-
total moles AgNO3 added = 0.4500 M x 60 ml = 27 mmol
AgNO3 + KSCN --> AgSCN + KNO3
moles KSCN added = 0.2400 M x 36.90 ml = 8.856 mmol
excess AgNO3 present = 8.856 mmol
actual AgNO3 reacted with AsO4^3- = 27 - 8.856 = 18.144 mmol
moles AsO4^3- present = 18.144/3 = 6.048 mmol
mass of AsO4^3- in original solution = 6.048 mmol x 138.92 g/mol/1000 = 0.840 g
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