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stepan [7]
2 years ago
14

Use the solubility rules from the Lab 4 introduction and your knowledge of qualitative separation schemes from the lab to answer

the following questions. The qualitative analysis experiment you did is actually an abbreviated version of a much larger analysis scheme in which many different cations are separated and identified. Suppose a mixture contains Ag , K , NH4 , Hg22 , Pb2 , Mg2 , Sr2 , Ba2 , Cu2 , Al3 and Fe3 .
(a) Which of the following ions could you separate, by causing them to precipitate, with the addition of HCl?
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
(b) After the addition of HCl, the above sample is centrifuged and decanted. Which of the following cations remaining in the supernatant could you separate, by causing them to precipitate, with the addition of H2SO4? (Hint: H2SO4 is a source of sulfate ions. Select all that apply.)
Ag+ K+ NH4+
Hg22+ Pb2+ Mg2+
Sr2+ Ba2+ Cu2+
Al3+ Fe3+
Chemistry
1 answer:
cluponka [151]2 years ago
4 0

Answer:

a13+a13

Explanation:

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The pH of the diluted HCl solution is 1.3.

Explanation:

Given:

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The pH of the diluted HCl solution.

Solution

  • The concentration of the HCl solution before dilution = M_1=8.0 M
  • The volume of the HCl solution taken for dilution = V_1=1.5 ml
  • The concentration of the HCl solution after dilution =M_2=?
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Using the Dilution equation:

M_1V_1=M_2V_2\\8.0M\times 1.5ml=M_2\times 250 mL\\M_2=\frac{8.0M\times 1.5ml}{250 mL}=0.048 M

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In the 1 M solution of HCl, there are 1 M of hydrogen ion, then the concentration of hydrogen ions in 0.048 M of HCl will be:

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The pH of the diluted HCl solution :

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The pH of the diluted HCl solution is 1.3.

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