Asked by Cherie
Suggest a range of sample masses for the indicated primary standard if it is desired to use between 35 and 45mL of titrant:
(a) 0.175 M HClO4 titrated against Na2CO3(CO2 product)
(b) 0.085 M HCl titrated against Na2C2O4
Na2C2O4-->Na2CO3+CO
CO3(2-)+2H(+)-->H2O+CO2
(c) 0.150 M NaOH titrated against benzoic acid
(d) 0.050 M Ba(OH)2 titrated against KH(IO3)2
(e) 0.075 M HClO4 titrated against TRIS
(f) 0.050 M H2SO4 titrated against Na2B4O7 * 10 H20
reaction:
B4O7(2-)+2 H30(+)+3H20-->4H3BO3
(a) 0.175 M HClO4 titrated against Na2CO3(CO2 product)
(b) 0.085 M HCl titrated against Na2C2O4
Na2C2O4-->Na2CO3+CO
CO3(2-)+2H(+)-->H2O+CO2
(c) 0.150 M NaOH titrated against benzoic acid
(d) 0.050 M Ba(OH)2 titrated against KH(IO3)2
(e) 0.075 M HClO4 titrated against TRIS
(f) 0.050 M H2SO4 titrated against Na2B4O7 * 10 H20
reaction:
B4O7(2-)+2 H30(+)+3H20-->4H3BO3
Answers
Answered by
DrBob222
a.
2HClO4 + Na2CO3 ==> 2NaCl + 2H2O
mols HClO4 = M x L = 0.035 x 0.175 = approx 0.006125
mols Na2CO3 = 1/2 that from 2 mol HClO4 = 1 mol Na2CO3
mol Na2CO3 = grams/molar mass or grams = mols x molar mass = 0.00306 x 106 = approx 0.320 g.
Now do the same thing with 45 mL instead of 35 mL to get the range. In practice you take approx the average.
2HClO4 + Na2CO3 ==> 2NaCl + 2H2O
mols HClO4 = M x L = 0.035 x 0.175 = approx 0.006125
mols Na2CO3 = 1/2 that from 2 mol HClO4 = 1 mol Na2CO3
mol Na2CO3 = grams/molar mass or grams = mols x molar mass = 0.00306 x 106 = approx 0.320 g.
Now do the same thing with 45 mL instead of 35 mL to get the range. In practice you take approx the average.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.