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3 years ago

Calculate the time needed for a constant current of 0.961 a to deposit 0.500 g of co(ii) as

Chemistry

1 answer:

Vera_Pavlovna [14]3 years ago

4 0

1.70 × 10³ seconds

<h3>Explanation </h3>

$\text{Co}^{2+}$ + 2 e⁻ → $\text{Co}$

It takes two moles of electrons to reduce one mole of cobalt (II) ions and deposit one mole of cobalt.

Cobalt has an atomic mass of 58.933 g/mol. 0.500 grams of Co contains $0.500 / 58.933 = 8.484\times 10^{-3} \; \text{mol}$ of Co atoms. It would take $2 \times 8.484 \times 10^{-3} = 0.01697 \; \text{mol}$ of electrons to reduce cobalt (II) ions and produce the $8.484\times 10^{-3} \; \text{mol}$ of cobalt atoms.

Refer to the Faraday's constant, each mole of electrons has a charge of around 96 485 columbs. The 0.01697 mol of electrons will have a charge of $1.637 \times 10^{3} \; \text{C}$ . A current of 0.961 A delivers 0.961 C of charge in one single second. It will take $1.637 \times 10^{3} / 0.961 = 1.70 \times 10^{3} \; \text{s}$ to transfer all these charge and deposit 0.500 g of Co.

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Answer:

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Explanation:

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Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH

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Here is the full question

Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (K a = 1.54 × 10 − 5), with 0.1000 M NaOH solution after the following additions of titrant (total volume of added base given):

a) 10.00 mL

pH =

b) 20.10 mL

pH =

c) 25.00 mL

pH =

Answer:

pH = 4.81

pH = 10.40

pH = 12.04

Explanation:

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $10.00 \ mL * \frac{L}{1000 \ mL }* \frac{0.1000 \ mol }{L}$

= 0.001000 mol

pKa of butanoic acid = - log Ka

= - log ( 1.54 × 10⁻⁵)

= 4.81

Equation for the reaction is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

The ICE Table is expressed as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.001000 0

Change - 0.001000 - 0.001000 + 0.001000

Equilibrium 0.001000 0 0.001000

Total Volume = (20.00 + 10.00 ) mL

= 30.00 mL = 0.03000 L

Concentration of [CH₃CH₂CH₂COOH] = $\frac{0.001000 \ mol}{ 0.03000 \ L }$

= 0.03333 M

Concentration of [CH₃CH₂COO⁻] = $\frac{0.001000 \ mol}{ 0.03000 \ L}$

= 0.03333 M

By Henderson- Hasselbalch equation

pH = pKa + log $\frac{conjugate \ base}{acid }$

pH = pKa + log $\frac{CH_3CH_2CH_2COO^-}{CH_3CH_2CH_2COOH}$

PH = 4.81 + log $\frac{0.03333}{0.03333}$

pH = 4.81

Thus; the pH of the resulted buffer solution after 10.00 mL of NaOH was added = 4.81

b )

After the equivalence point, we all know that the pH of the solution will now definitely be determined by the excess H⁺

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $20.10 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002010 mol

Following the previous equation of reaction , The ICE Table for this process is as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002010 0

Change - 0.002000 -0.002000 + 0.002000

Equilibrium 0 0.000010 0.002000

We can see here that the base is present in excess;

NOW, number of moles of base present in excess

= ( 0.002010 - 0.002000) mol

= 0.000010 mol

Total Volume = (20.00 + 20.10 ) mL

= 40.10 mL × $\frac{1 \ L}{1000 \ mL }$

= 0.04010 L

Concentration of acid [OH⁻] = $\frac{0.000010 \ mol}{0.04010 \ L }$

= $2.494*10^{-4} M$

Using the ionic product of water:

$[H_3O^+] = \frac{K \omega }{[OH^-]}$

where

$K \omega = 10^{-14}$

$[H_3O^+] = \frac{1.0*10^{-14}}{2.494*10^{-14}}$

= $4.0*10^{-11}M$

pH = - log $[H_3O^+}]$

pH = - log [ $4.0*10^{-11}M$ ]

pH = 10.40

Thus, the pH of the solution after the equivalence point = 10.40

After the equivalence point, pH of the solution is determined by the excess H⁺.

Number of moles of butanoic acid

= $20.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002000 mol

Number of moles of NaOH added

= $25.00 \ mL * \frac{L}{1000 \ mL} * \frac{0.1000 \ mol}{ L}$

= 0.002500 mol

From our chemical equation; The ICE Table can be illustrated as follows:

CH₃CH₂CH₂COOH + OH⁻ -----> CH₃CH₂COO⁻ + H₂O

Initial 0.002000 0.002500 0

Change - 0.002000 -0.002000 +0.002000

Equilibrium 0 0.000500 0.002000

Base is present in excess

Number of moles of base present in excess = [ 0.002500 - 0.002000] mol

= 0.000500 mol

Total Volume = ( 20.00 + 25.00 ) mL

= 45.00 mL

= 45.00 × $\frac{1 \ L}{1000 \ mL }$

= 0.04500 L

Concentration of acid [OH⁻] = $\frac{0.0005000 \ mol}{ 0.04500 \ L }$

= 0.01111 M

Using the ionic product of water $[H_3O^+] = \frac{K \omega }{[OH^+]}$

= $\frac{1.0*10^{-14}}{0.01111}$

= $9.0*10^{-13}$ M

pH = - log $[H_3O^+}]$

pH = - log [ $9.0*10^{-13}M$ ]

pH = 12.04

Thus, the pH of the solution after the equivalence point = 12.04

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lorasvet [3.4K]

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Explanation :

In this case the heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of iron metal = $0.45J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of iron metal = 32.3 g

$m_2$ = mass of water = ?

$T_f$ = final temperature of mixture = $59.2^oC$

$T_1$ = initial temperature of iron metal = $21.9^oC$

$T_2$ = initial temperature of water = $63.5^oC$

Now put all the given values in the above formula, we get

$32.3g\times 0.45J/g^oC\times (59.2-21.9)^oC=-m_2\times 4.18J/g^oC\times (59.2-63.5)^oC$

$m_2=30.16g\approx 3.0\times 10^1g$

Therefore, the mass of the water in two significant figures is, $3.0\times 10^1g$

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