What force pulls electrons and protons together?

Calculate the pH of each of the following solutions. a. 0.100 M propanoic acid (HC3H5O2, Ka 1.3 105 ) b. 0.100 M sodium propanoa

Allisa [31]

Answer:

a) pH = 2.95

b) pH = 8.94

c) pH = 7.00

d) pH = 4.89

Explanation:

a) The reaction is:

CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)(1)

At equlibrium: 0.1 - x x x

The dissociation constant of equation (1), Ka, is:

$K_{a} = \frac{[C_{3}H_{5}O_{2}^{-}][H_{3}O^{+}]}{[C_{3}H_{6}O_{2}]}$

$1.3\cdot 10^{-5} = \frac{x^{2}}{(0.1 - x)}$ (2)

Solving equation (2) for x, we have:

x = 0.00113 = [H₃O⁺] = [CH₃CH₂COO⁻]

Hence, the pH is:

$pH = -log [H_{3}O^{+}] = -log (0.00113) = 2.95$

b) The reaction is:

CH₃CH₂COO⁻(aq) + H₂O(l) ⇄ CH₃CH₂COOH(aq) + OH⁻(aq)

At equlibrium: 0.1 - x x x

The dissociation constant of the above equation, Kb, is:

$K_{b} = \frac{[C_{3}H_{6}O_{2}][OH^{-}]}{[C_{3}H_{5}O_{2}^{-}]}$

Also, we have that:

$K_{w} = K_{a}*K_{b} \rightarrow K_{b} = \frac{K_{w}}{K_{a}} = \frac{1 \cdot 10^{-14}}{1.3 \cdot 10^{-5}} = 7.69 \cdot 10^{-10}$

$7.69\cdot 10^{-10} = \frac{x^{2}}{0.1 - x}$ (3)

Solving equation (3) for x, we have:

x = 8.77x10⁻⁶ = [OH⁻] = [CH₃CH₂COOH]

Hence, the pH is:

$pOH = -log[OH^{-}] = -log(8.77\cdot 10^{-6}) = 5.06 \rightarrow pH = 14 - pOH = 8.94$

c) Pure water:

H₂O ⇄ H⁺ + OH⁻

$K_{w} = [H^{+}][OH^{-}] \rightarrow 1\cdot 10^{-14} = [H^{+}][OH^{-}]$

Since is pure water: [H⁺] = [OH⁻]

$1\cdot 10^{-14} = [H^{+}]^{2} \rightarrow [H^{+}] = \sqrt{1\cdot 10^{-14}} = 1 \cdot 10^{-7}$

Therefore, the pH is:

$pH = -log [H^{+}] = -log (1 \cdot 10^{-7}) = 7.00$

d) The reaction is:

CH₃CH₂COOH(aq) + H₂O(l) ⇄ CH₃CH₂COO⁻(aq) + H₃O⁺(aq)

At equlibrium: 0.1 0.1 x

The pH is:

$pH = pKa + log(\frac{[C_{3}H_{5}O_{2}^{-}]}{[C_{3}H_{6}O_{2}]}) = -log(1.3 \cdot 10^{-5}) + log(\frac{0.1}{0.1}) = 4.89$

I hope it helps you!

4 0

2 years ago

he thermite reaction, in which powdered aluminum reacts with iron(III) oxide, is highly exothermic. 2 Al(s) + Fe2O3(s) Al2O3(s)

Afina-wow [57]

The given question is incomplete. The complete question is

The thermite reaction, in which powdered aluminum reacts with iron oxide, is highly exothermic: $2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)$ . Use standard enthalpies of formation to find ΔH∘rxn for the thermite reaction. Express the heat of the reaction in kilojoules to four significant figures.

Answer: ΔH∘rxn for the thermite reaction is -851.5 kJ

Explanation:

The balanced chemical reaction is :

$2Al(s)+Fe_2O_3(s)\rightarrow Al_2O_3(s)+2Fe(s)$

We have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{Fe}\times \Delta H_f^0_{(Fe)}+n_{Al_2O_3}\times \Delta H_f^0_{(Al_2O_3)}]-[n_{Al}\times \Delta H_f^0_(Al)+n_{Fe_2O_3}\times \Delta H_f^0_{(Fe_2O_3)}]$

where,

We are given:

$\Delta H^o_f_{(Fe_2O_3(s))}=-824.2kJ/mol\\\Delta H^o_f_{(Al_2O_3(s))}=-1675.7kJ/mol\\\Delta H^o_f_{(Fe(s))}=0kJ/mol\\\Delta H^o_f_{(Al(s)}=0kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(2\times 0)+(1\times -1675.5)]-[(2\times 0)+(1\times -824.2)]=-851.5kJ$

ΔH∘rxn for the thermite reaction is -851.5 kJ

4 0

2 years ago

5. Molds and bacteria contain hydrolytic enzymes used to breakdown proteins,

Tpy6a [65]

Answer:Temperature affects storage time and food deteriorates faster at higher temperatures. Microorganisms grow rapidly at room temperature . Thus to slow microbial growth, enzymatic and oxidative processes, food must be stored at room temperature.

Due to decomposition reactions with oxygen or carbon dioxide in the air, meat begins to feel slimy and smell spoiled. ... Because of its smaller surface area, the unsliced meat would have less collisions and therefore, a slower reaction rate.

Explanation:

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AleksAgata [21]

Butterflies like to drink the sweet nectar from flowers.

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