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blagie [28]
2 years ago
13

Hardness in groundwater is due to the presence of metal ions, primarily Mg2+ and Ca2+ . Hardness is generally reported as ppm Ca

CO3 . To measure water hardness, a sample of groundwater is titrated with EDTA, a chelating agent, in the presence of the indicator Eriochrome Black T, symbolized as In. Eriochrome Black T, a weaker chelating agent than EDTA, is red in the presence of Ca2+ and turns blue when Ca2+ is removed. red blue Ca(In)2+ + EDTA ⟶ Ca(EDTA)2+ + In A 50.00 mL sample of groundwater is titrated with 0.0450 M EDTA . If 13.70 mL of EDTA is required to titrate the 50.00 mL sample, what is the hardness of the groundwater in molarity and in parts per million of CaCO3 by mass? Assume that Ca2+ accounts for all of the hardness in the groundwater.
Chemistry
1 answer:
JulsSmile [24]2 years ago
7 0

Answer:

[Ca⁺²] = 0.0123 M

1234 ppm of CaCO₃

Explanation:

Ca(In)²⁺ + EDTA ⟶ Ca(EDTA)²⁺ + In

To solve this problem we need to keep in mind the definition of Molarity (moles/L) and parts per million (mg CaCO₃ / L water). First we <u>calculate the moles of EDTA</u>:

  • 0.0450 M * 13.70 mL = 0.6165 mmol EDTA

Looking at the reaction we see that one mol of EDTA reacts with one mol of Ca(In)²⁺, so we have as well 0.6165 mmol Ca(In)²⁺ (which for this case is the same as Ca⁺²). We calculate the molarity of Ca⁺²:

  • [Ca⁺²] = 0.6165 mmol / 50 mL = 0.0123 M

For calculating the concentration in ppm, we use the moles of Ca⁺² and the molecular weight of CaCO₃ (100.09 g/mol):

  • 0.6165 mmol Ca⁺² * \frac{1mmolCaCO_{3}}{1mmolCa^{+2}} * 100.09 mg/mmol = 61.70 mg CaCO₃

<u>That is the mass of CaCO₃ present in 50 mL</u> (or 0.05 L) of water, so the concentration in ppm is:

  • 61.70 mg / 0.05 L = 1234 ppm
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Answer: the equilibrium will be displaced to the right leading an increase on the quantities of y(g) and z(s).

Justification:

According to the rules of equilibrium, based on Le Chatellier's priciple, any change in a system in equilibrium will be tried to be compensated to restablish the equilibrium

The higher the amount, and so the concentration, of X(g), the more will the forward reaction proceed leading to an increase on the concentration of the products y(g) and z (s). Look that that will also be accompanied by a decreasing on the pressure, since 2 molecules of the gas X(g) are converted into 1 molecule of the gas y(g).
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2 years ago
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Which of the following statements concerning a metal crystallized in a face-centered cubic cell is/are CORRECT? 1. One metal ato
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Answer:

Only 3 is correct.

Explanation:

The crystal of a metal or an ionic compound is called a cell, and there are 7 types of unit cells: cubic, tetragonal, orthorhombic, monoclinic, hexagonal, rhombohedral, and triclinic.

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2 years ago
Determine whether or not the mixing of each of the two solutions indicated below will result in a buffer.
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Part A

75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF

This combination will form a buffer.

Explanation

Here, weak acid HF and its conjugate base F- is available in the solution

Part B

150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl

This combination cannot form a buffer.

Explanation

Here, moles of HF = 0.15 x 0.1 = 0.015 moles

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Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution

Part C

165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH

This combination will form a buffer.

Explanation

Moles of HF = 0.165 x 0.1 = 0.0165 moles

Moles of KOH = 0.135 x 0.05 = 0.00675 moles

Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer

Part D

125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl

This combination will form a buffer

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Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer

Part E

105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl

This combination will form a buffer

Explanation

Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles

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Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
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Answer:

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Explanation:

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Equilibrium constant of the reaction :

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After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

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