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johnportugal22 johnportugal22
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10 years ago
1) A 6.00-mL sample of a sulfuric acid solution of unknown concentration is titrated with a 0.1513 M Sodium Hydroxide solution. A volume of 4.99 mL of the base was required to reach the endpoint. What is the concentration of the unknown acid solution? Write the neutralization reaction, balance it, name the products THEN Solve the problem above.


2) A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine:

 C6H8O6(aq) + I2(aq) →    C6H6O6(aq) + HI(aq)
(a) A 25.0 mL sample of juice requires 5.8 mL of a 0.0364 M  I2 solution for reaction. how many moles of ascorbic acid are in the sample?
(b) What is the molarity of the acid?
(c) If a person wanted to consume the FDA recommended 0.060 g of Vitamin C, how many ounces of juice would that person need to consume? (4 qts = 128 fluid oounces; 1L = 1.057 qt)
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wrote...
10 years ago
1) A 6.00-mL sample of a sulfuric acid solution of unknown concentration is titrated with a 0.1513 M Sodium Hydroxide solution. A volume of 4.99 mL of the base was required to reach the endpoint. What is the concentration of the unknown acid solution? Write the neutralization reaction, balance it, name the products THEN Solve the problem above.

no of milli eq are same..
M1v1=M2v2
5.99*0.1401=5*M
0.839199=5M
M=0.1678
2 NAOH + H2So4 ---------- NaSo4 + 2H20
sodium hydroxide + sulphuric acid ------------ sodium sulphate + water

2) A test for vitamin C (ascorbic acid, C6H8O6) is based on the reaction of the vitamin with iodine:

 C6H8O6(aq) + I2(aq) →    C6H6O6(aq) + HI(aq)
(a) A 25.0 mL sample of juice requires 5.8 mL of a 0.0364 M  I2 solution for reaction. how many moles of ascorbic acid are in the sample?
(b) What is the molarity of the acid?
(c) If a person wanted to consume the FDA recommended 0.060 g of Vitamin C, how many ounces of juice would that person need to consume? (4 qts = 128 fluid oounces; 1L = 1.057 qt)

a) moles of ascorbic acid = moles of I2

= (5.8*0.0364)/1000 (Volume should be in L) and n = M*V

= 2.1*10^-4 moles or 0.21 mili moles



b) Molarity = n/V = (2.1*10^-4) / (25/1000) = 0.0084
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