CHEMISTRY COLLEGE

Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C. How are these numbers affected by the addition of 0.1 mol/dm3 of KCL? At what distance from the particle surface (r) has the potential decayed to 1% of its initial value?

Answers

Answer 1

Answer:

Explanation:

The given data is as follows.

Concentration = 0.1 $mol/dm^(3)$

= 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

= $6.022 * 10^(25) ions/m^(3)$

T = $30^(o)C$ = (30 + 273) K = 303 K

Formula for electric double layer thickness ( $\lambda_(D)$ ) is as follows.

$\lambda_(D)$ = $(1)/(k) = \sqrt (\varepsilon \varepsilon_(o) K_(g)T)/(2 n^(o) z^(2) \varepsilon^(2))$

where, $n^(o)$ = concentration = $6.022 * 10^(25) ions/m^(3)$

Hence, putting the given values into the above equation as follows.

$\lambda_(D)$ = $\sqrt (\varepsilon \varepsilon_(o) K_(g)T)/(2 n^(o) z^(2) \varepsilon^(2))$

= $\sqrt (78 * 8.854 * 10^(-12) c^(2)/Jm * 1.38 * 10^(-23)J/K * 303 K)/(2 * 6.022 * 10^(25) ions/m^(3) * (1)^(2) * (1.6 * 10^(-19)C)^(2))$

= $9.669 * 10^(-10)$ m

or, = $9.7 A^(o)$

= 1 nm (approx)

Also, it is known that $\lambda_(D)$ = $\sqrt (1)/(n^(o))$

Hence, we can conclude that addition of 0.1 $mol/dm^(3)$ of KCl in 0.1 $mol/dm^(3)$ of NaBr " $\lambda_(D)$ " will decrease but not significantly.

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Consider the titration of a 73.9 mL sample of 0.13 M HC2H3O2 with 6.978 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the initial pH before any NaOH is added. Express your answer using two decimal places.Consider the titration of a 46.6 mL sample of 0.078 M HC2H3O2 with 1.135 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the volume of added base required to reach the equivalence point. Answer in units of milliliters.

Consider the titration of a 17.2 mL sample of 0.128 M HC2H3O2 with 0.155 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the pH at 0.46 mL of added base.

Answers

Answer:

1. pH = 2,82

2. 3,20mL of 1,135M NaOH

3. pH = 3,25

Explanation:

The buffer of acetic acid (HC₂H₃O₂) is:

HC₂H₃O₂ ⇄ H⁺ + C₂H₃O₂⁻

The reaction of HC₂H₃O₂ with NaOH produce:

HC₂H₃O₂ + NaOH → C₂H₃O₂⁻ + Na⁺ + H₂O

And ka is defined as:

ka = [H⁺] [C₂H₃O₂⁻] / [HC₂H₃O₂] = 1,8x10⁻⁵ (1)

1. When in the solution you have just 0,13M HC₂H₃O₂ the concentrations in equilibrium will be:

[H⁺] = x

[C₂H₃O₂⁻] = x

[HC₂H₃O₂] = 0,13 - x

Replacing in (1)

[x] [x] / [0,13-x] = 1,8x10⁻⁵

x² = 2,34x10⁻⁶ - 1,8x10⁻⁵x

x² - 2,34x10⁻⁶ + 1,8x10⁻⁵x = 0

Solving for x:

x = - 0,0015 (Wrong answer, there is no negative concentrations)

x = 0,0015

As [H⁺] = x = 0,0015 and pH is -log [H⁺], pH of the solution is 2,82

2. The equivalence point is reached when moles of HC₂H₃O₂ are equal to moles of NaOH. Moles of HC₂H₃O₂ are:

0,0466L × (0,078mol / L) = 3,63x10⁻³ moles of HC₂H₃O₂

In a 1,135M NaOH, these moles are reached with the addition of:

3,63x10⁻³ moles × (L / 1,135mol) = 3,20x10⁻³L = 3,20mL of 1,135M NaOH

3. The initial moles of HC₂H₃O₂ are:

0,0172L × (0,128mol / L) = 2,20x10⁻³ moles of HC₂H₃O₂

As the addition of NaOH spent HC₂H₃O₂ producing C₂H₃O₂⁻. Moles of C₂H₃O₂⁻ are equal to moles of NaOH and moles of HC₂H₃O₂ are initial moles - moles of NaOH. That means:

0,46x10⁻³L NaOH × (0,155mol / L) = 7,13x10⁻⁵ moles of NaOH ≡ moles of C₂H₃O₂⁻

Final moles of HC₂H₃O₂ are:

2,20x10⁻³ - 7,13x10⁻⁵ = 2,2187x10⁻³ moles of HC₂H₃O₂

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [C₂H₃O₂⁻] / [HC₂H₃O₂]

Where pka is -log ka = 4,74. Replacing:

pH = 4,74 + log₁₀ [7,13x10⁻⁵] / [2,2187x10⁻³ ]

pH = 3,25

I hope it helps!

Refer to the following balance equation : 2c2h6 + 7o2 = 4co2 + 6 h20. How many moles of c2h6 will combust completely with 2.0 moles of oxygen gas. How many grams of water will be produced when 30.0 g of c2h6 reacts completely with oxygen?

Answers

1. From the balanced equation given above, the ratio of the number of moles of the hydrocarbon and oxygen is equal to 2/7. Given that there are 2 moles of oxygen,

moles hydrocarbon = (2 moles O2)(2 moles HC / 7 moles oxygen)

Simplifying,
moles HC = 4/7 or 0.57 moles HC

Answer: 0.57 moles HC

2. The calculation for the mass of water is shown below with the dimensional analysis and conversion factors,

(30 g C2H6)1 mole C2H6/30 g C2H6)(6 molesH2O/2 moles C2H6)(18 g H2O/1 mole C2H6)

Simplifying,
mass = 54 grams of water

6CO2 + 6H20 --> C6H12O6 + 602What is the total number of moles of CO2 needed to make 2 moles of CH1206?

Answers

Answer:

12 mol CO₂

General Formulas and Concepts:

Atomic Structure

Compounds
Moles
Mole Ratio

Stoichiometry

Analyzing reactions rxn
Using Dimensional Analysis

Explanation:

Step 1: Define

Identify

[rxn] 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

[Given] 2 mol C₆H₁₂O₆

[Solve] mol CO₂

Step 2: Identify Conversions

[rxn] 6CO₂ → C₆H₁₂O₆

Step 3: Convert

[DA] Set up: $\displaystyle 2 \ mol \ C_6H_(12)O_6((6 \ mol \ CO_2)/(1 \ mol \ C_6H_(12)O_6))$
[DA] Multiply [Cancel out units]: $\displaystyle 12 \ mol \ CO_2$

Solve for x . 875 = 5 x 3 Express the answer to the hundredths place (i.e., two digits after the decimal point).

Answers

Answer : The value of 'x' for this expression is, 5.59

Explanation :

The given expression is:

$875=5x^3$

Now we have to determine the value of 'x' by solving the above expression.

$875=5x^3$

$(875)/(5)=x^3$

$175=x^3$

$x=(175)^(1/3)$

$x=5.59$

Thus, the value of 'x' for this expression is, 5.59

Glucose, C 6 H 12 O 6 , is used as an energy source by the human body. The overall reaction in the body is described by the equation C 6 H 12 O 6 ( aq ) + 6 O 2 ( g ) ⟶ 6 CO 2 ( g ) + 6 H 2 O ( l ) Calculate the number of grams of oxygen required to convert 58.0 g of glucose to CO 2 and H 2 O . mass of O 2 : 61.76 g Calculate the number of grams of CO 2 produced.