CHEMISTRY COLLEGE

Calculate the solubilities of the following compounds in a 0.02 M solution of barium nitrate using molar concentrations, first ignoring ionic strength and activities. a. silver iodate
b. barium sulfate
c. Repeat the above calculations using ionic strength and activities.

Answers

Answer 1
Answer:

Answer:

a. 1.7 × 10⁻⁴ mol·L⁻¹; b. 5.5 × 10⁻⁹ mol·L⁻¹

c. 2.3 × 10⁻⁴ mol·L⁻¹;    5.5 × 10⁻⁸ mol·L⁻¹

Explanation:

a. Silver iodate

Let s = the molar solubility.  

                     AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq); Ksp = 3.0 × 10⁻⁸

E/mol·L⁻¹:                               s               s

K_(sp) =\text{[Ag$^(+)$][IO$_(3)$$^(-)$]} = s* s =  s^(2) = 3.0* 10^(-8)\ns = \sqrt{3.0* 10^(-8)} \text{ mol/L} = 1.7 * 10^(-4) \text{ mol/L}

b. Barium sulfate

                     BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq); Ksp = 1.1 × 10⁻¹⁰

I/mol·L⁻¹:                                0.02             0

C/mol·L⁻¹:                                 +s              +s

E/mol·L⁻¹:                            0.02 + s          s

K_(sp) =\text{[Ba$^(2+)$][SO$_(4)$$^(2-)$]} = (0.02 + s) * s \approx  0.02s = 1.1* 10^(-10)\ns = (1.1* 10^(-10))/(0.02) \text{ mol/L} = 5.5 * 10^(-9) \text{ mol/L}

c. Using ionic strength and activities

(i) Calculate the ionic strength of 0.02 mol·L⁻¹ Ba(NO₃)₂

The formula for ionic strength is  

\mu = (1)/(2) \sum_(i) {c_(i)z_(i)^(2)}\n\n\mu = (1)/(2) (\text{[Ba$^(2+)$]}\cdot (2+)^(2) + \text{[NO$_(3)$$^(-)$]}*(-1)^(2)) = (1)/(2) (\text{0.02}* 4 + \text{0.04}*1)= (1)/(2) (0.08 + 0.04)\n\n= (1)/(2) *0.12 = 0.06

(ii) Silver iodate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^(2)√(I) = -0.051(1)^(2)√(0.06) = -0.51* 0.24 = -0.12\n\gamma = 10^(-0.12) = 0.75

b. Calculate the solubility

AgIO₃(s) ⇌ Ag⁺(aq) + IO₃⁻(aq)

K_(sp) =\text{[Ag$^(+)$]$\gamma_(Ag^(+))$[IO$_(3)$$^(-)$]$\gamma_{IO_(3)^(-)}$} = s*0.75* s * 0.75 =0.56s^(2)= 3.0 * 10^(-8)\ns^(2) = (3.0 * 10^(-8))/(0.56) = 5.3 * 10^(-8)\n\ns =2.3 * 10^(-4)\text{ mol/L}

(iii) Barium sulfate

a. Calculate the activity coefficients of the ions

\log \gamma = -0.51z^(2)√(I) = -0.051(2)^(2)√(0.06) = -0.51*16* 0.24 = -0.50\n\gamma = 10^(-0.50) = 0.32

b. Calculate the solubility

BaSO₄(s) ⇌ Ba²⁺(aq) + SO₄²⁻(aq

K_(sp) =\text{[Ba$^(2+)$]$\gamma_( Ba^(2+))$[SO$_(4)$$^(2-)$]$\gamma_{ SO_(4)^(2-)}$} = (0.02 + s) * 0.32* s* 0.32 \approx  0.02*0.10s\n2.0* 10^(-3)s = 1.1 * 10^(-10)\ns = (1.1* 10^(-10))/(2.0 * 10^(-3)) \text{ mol/L} = 5.5 * 10^(-8) \text{ mol/L}

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A student dissolves 12.g of sucrose C12H22O11 in 300.mL of a solvent with a density of 1.01/gmL . The student notices that the volume of the solvent does not change when the sucrose dissolves in it. Calculate the molarity and molality of the student's solution. Round both of your answers to 2 significant digits. molarity = molality =

Answers

Answer:

Molarity → 0.17 M

Molality → 0.11 m

Explanation:

The student notices that the volume of the solvent does not change when the sucrose dissolves in it; therefore we assume the volume of solvent as solution.

Molarity = Mol of solute/L

Let's calculate the mol of solute (mass / molar mass)

12 g / 342 g/mol = 0.0351 moles

Let's conver the volume (mL) to L

300 mL / 1000 = 0.3 L

Molarity (mol/L) = 0.0351 mol / 0.3L → 0.17 M

Molality = mol of solute / 1kg of solvent.

Let's find out the mass of solvent with the density

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1.01 g/mL = Solvent mass / 300 mL

1.01 g/mL . 300 mL = Solvent mass →303 g

Let's convert the mass to kg

303 g / 1000 = 0.303 kg

Molality (mol/kg) → 0.0351 mol / 0.303kg  = 0.11 m

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B.) Positively charged
C.) Strongly ionic
D.) Negatively charged

Answers

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Which of the following combinations represents an element with a net charge of +1 with a mass number of 75?a) 35 (o), 35 (+), 34 (-)b) 40 (o), 40 (+), 39 (-)c) 40 (o), 35 (+), 34 (-)d) 37 (o), 38 (+), 34 (-)e) 40 (o), 35 (+), 35 (-)

Answers

Answer:

c) 40 (o), 35 (+), 34 (-)

Explanation:

Let us represent the element with P

Given information about the element:

Charge on P = +1

Mass number of P = 75

We can express the atom as ⁴⁵P¹⁺

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The mass number 75 is the number of protons plus neutrons.

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Protons = 35

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Draw a well-labelled diagram showing how your body digests food​

Answers

i too used it

i thought it will help

nice time. .....

Explanation:

...................

A solution of HNO3HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 2H++Na2CO3⟶2Na++H2O+CO2 A volume of 27.71±0.05 mL27.71±0.05 mL of HNO3HNO3 solution was required for complete reaction with 0.9585±0.0007 g0.9585±0.0007 g of Na2CO3Na2CO3 , (FM 105.988±0.001 g/mol105.988±0.001 g/mol ). Find the molarity of the HNO3HNO3 solution and its absolute uncertainty.

Answers

Answer:

(0,653±0,002) M of HNO₃

Explanation:

The reaction of standarization of HNO₃ with Na₂CO₃ is:

2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻

To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (27,71±0,05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus:

0,9585 g of Na₂CO₃ × ( 1 mole / 105,988 g) =

9,043×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,809×10⁻² moles of HNO₃

Molarity is moles divide liters, thus, molarity of HNO₃ is:

1,809×10⁻² moles / 0,02771 L = 0,6527 M of HNO₃

The absolute uncertainty of multiplication is the sum of relative uncertainty, thus:

ΔM = 0,6527M× (0,0007/0,9585 + 0,001/105,988 + 0,05/27,71) =

0,6527 M× 2,54×10⁻³ = 1,7×10⁻³ M

Thus, molarity of HNO₃ solution and its absolute uncertainty is:

(0,653±0,002) M of HNO₃

I hope it helps!
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