21 and 22 please 21.A phosphate buffer solution (25.00 ml sample) used for a gro
ID: 515171 • Letter: 2
Question
21 and 22 please
21.A phosphate buffer solution (25.00 ml sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The components of the buffer were sodium monohydrogen phosphate and sodium of dihydrogen phosphate. the first endpoint occured at a volume of 10.32mL, and the second occured after an additional 18.62mL was added, for a total volume of 28.94mL. what was the total concentration of phosphate(in any form) in the buffer?
a.0.03992M
b. 0.01198M
c.0.04243M
d.0.07448M
e.0.08382M
22. vitamin C is monoprotic weak acid, which alsois called ascorbic acid(C6H8O6, 176g/mol). a vitamin C tablet weighing 0.75g was dissolved in 50.0mL of water and titrated with 0.250M sodium hydroxide. it took 12.5mL of the NaOH solution to reach the end point. what is the percentage of vitamin C in the tablet? only the vitamin C reacted with the NaOH.
a.98.7%
b.64.2%
c.87.7%
d.95.5%
e.73.3%
A phosphate buffer solution (25.00 ml sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The components of the buffer were sodium monohydrogen phosphate and sodium dihydrogen phosphate. The first endpoint occurred at a volume of 10.32 mL, and the second occurred after an additional 18.62 mL, was added, for a total volume of 28, 94 mL. What was the total concentration of phosphate (in any form) in the buffer? a. 0.03992 M b. 0.1198 M c. 0.04243 M d. 0.07448 M e. 0.08382 M A vitamin C is a monoprotic weak acid, which also is called ascorbic acid (C H_ O_, 176 g/mol). A vitamin C tablet weighing 0.75 g was dissolved in 50.0 mL of water and titrated with 0.250 M sodium hydroxide, It took 12.5 mL of the NaOH solution to reach the end point. What is the percentage of vitamin C in the tablet? Only the vitamin C reacted with the NaOH. a. 98.7% b. 64.2% c. 87.7% d. 95.5% e. 73.3%Explanation / Answer
21.A phosphate buffer solution (25.00 ml sample) used for a growth medium was titrated with 0.1000 M hydrochloric acid. The components of the buffer were sodium monohydrogen phosphate and sodium of dihydrogen phosphate. the first endpoint occured at a volume of 10.32mL, and the second occured after an additional 18.62mL was added, for a total volume of 28.94mL. what was the total concentration of phosphate(in any form) in the buffer?
a.0.03992M
b. 0.01198M
c.0.04243M
d.0.07448M-----------answer
e.0.08382M
endpoint occurred at a volume of 10.32 mL
reacted with the sodium monohydrogenphosphate
1 mol (HPO4)^-2 & 1 mol HCl --> 1 mol (H2PO4)^-1
use molarity to find the moles HCl in the 10.32 mL, (0.01032 Litres):
(0.01032 Litres) (0.1000 mol / Litre) = 0.001032 mol of HCl used
that 0.001032 mol of HCl used reacted by the equation:
1 mol (HPO4)^-2 & 1 mol HCl --> 1 mol (H2PO4)^-1
with an equal number of moles of sodium monohydrogenphosphate (HPO4)^-2
so, you had 0.001032 mol of sodium monohydrogenphosphate (HPO4)^-2
however, by the equation:
1 mol (HPO4)^-2 & 1 mol HCl --> 1 mol (H2PO4)^-1
it produces an equal number of the sodium mdihydrogenphosphate (H2PO4)^-1
which will use another 10.32 ml of HCl to get to the second end point:
1 mol (H2PO4)^-1 & 1 mol HCl --> 1 mol H3PO4
for a total of 20.64 mL of HCl
the remainder of the HCl, (28.94 mL - 20.64 mL = 8.30 ml)), was used to take the buffer's original
sodium dihydrogenphosphate (H2PO4)^-1 from the first end point to the second:
1 mol (H2PO4)^-1 & 1 mol HCl --> 1 mol H3PO4
find moles of HCl required:
(0.00830 Litres of HCl) (0.1000 mol / Litre) = 0.000830 moles of HCl required
by the equation:
1 mol (H2PO4)^-1 & 1 mol HCl --> 1 mol H3PO4
the 0.000830 moles of HCl required reacted with an equal 0.000830 moles of (H2PO4)^-1
find the total moles of phosphate (in any form) in the buffer:
(0.001032 mol of sodium monohydrogenphosphate)
&
(0.000830 mol of sodium dihydrogenphosphate)
totals
0.001862 mol of phosphate
total concentration of phosphate (in any form) in the 25.00 ml sample of buffer:
=(0.001862 mol of phosphate) / (0.02500 Litres) = 0.07448 Molar phosphate buffer
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vitamin C is monoprotic weak acid, which alsois called ascorbic acid(C6H8O6, 176g/mol). a vitamin C tablet weighing 0.75g was dissolved in 50.0mL of water and titrated with 0.250M sodium hydroxide. it took 12.5mL of the NaOH solution to reach the end point. what is the percentage of vitamin C in the tablet? only the vitamin C reacted with the NaOH.
a.98.7%
b.64.2%
c.87.7%
d.95.5%
e.73.3%--answer
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