1. If 14.95 mL of a 0.1250 M HCl solution is added to reach the endpoint of a ti
ID: 1048860 • Letter: 1
Question
1. If 14.95 mL of a 0.1250 M HCl solution is added to reach the endpoint of a titration, how many moles of HCl have been added?
0.001869 moles
0.008361 moles
8.361 moles
0.1196 moles
1.869 moles
QUESTION 2
A buret is read to how many decimal places?
3
can only be read to the ones place - so no decimal places
2
4
1
QUESTION 3
As the endpoint nears what happens to the color?
nothing can visibly be seen until the endpoint happens
the color begins being more prominent, but still disappears
the color is very vibrant but then will go away when the endpoint happens
there is no color change yet
color is perminent
QUESTION 4
A 20.00 mL sample of 0.1015 M nitric acid is introduced into a flask, and water is added until the volume of the solution reaches 250. mL. What is the concentration of nitric acid in the final solution?
5.08 × 10–4 M
3.25 × 10–2 M
8.12 × 10–3 M
0.406 M
1.27 M
QUESTION 5
If 25.0 mL of an unknown concentration of sodium hydroxide is added to a flask and then titrated with 0.1025 M HCl. If 24.69 mL of the acid was added, what is the concentration of sodium hydroxide?
63.27 M
0.6741 M
0.1012 M
0.1038 M
9.879 M
0.001869 moles
0.008361 moles
8.361 moles
0.1196 moles
1.869 moles
Explanation / Answer
Ans 1: Correct option is 0.001869 moles
Explanation:
Number of moles of HCl = Volume of HCl (L) x Molarity of HCl solution
Number of moles of HCl = (14.95mL x 1L/1000mL) x 0.1250 M
Number of moles of HCl = 0.001869 mol
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Ans 2: Correct answer is 2.
Explanation:
The smallest gradation on burette is 0.1mL. However, second decimal can be determined depending upon where the bottom of miniscus lies between two 0.1 marks. Second decimal will be an estimated value.
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Ans 3: correct answer is the color begins being more prominent, but still disappears
Explanation:
As the end point nears, the color starts becoming more prominent but it disappears. At the end point, the color does not disappear.
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Ans 4: Correct answer is 8.12 × 10–3 M
Explanation:
Number of moles of nitric acid = Volume (L) x Molarity
Number of moles of nitric acid = (20.00mL x 1L/1000mL) x 0.1015
Number of moles of nitric acid = 0.00203 mol
Thus, molarity of diluted solution = number of moles / volume(L)
molarity of diluted solution = (0.00203 mol) / 0.250L
molarity of diluted solution = 8.12 × 10–3 M
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Ans 5: Correct answer is 0.1012 M
Explanation:
Moles of HCl required = Volume of HCl (L) x molarity of HCl
Moles of HCl required = (24.69mL x 1L/1000mL) x 0.1025M
Moles of HCl required = 0.002531 mol
Since, at the end point moles of HCl = moles of NaOH
Therefore, moles of NaOH = 0.002531 mol
Concentration of NaOH = Moles / volume (L)
Concentration of NaOH = 0.002531 mol / 0.025L
Concentration of NaOH = 0.1012M
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