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3 years ago

HC2H3O2 (aq) + H2O (l) ⇔ C2H3O2- (aq) + H3O+ (aq) Ka = 1.8 x 10-5

Chemistry

1 answer:

marin [14]3 years ago

5 0

The concentration of [H3O⁺]=2.86 x 10⁻⁶ M

<h3>Further explanation</h3>

In general, the weak acid ionization reaction

HA (aq) ---> H⁺ (aq) + A⁻ (aq)

Ka's value

$\large {\boxed {\bold {Ka \: = \: \frac {[H ^ +] [A ^ -]} {[HA]}}}}$

Reaction

HC₂H₃O₂ (aq) + H₂O (l) ⇔ (aq) + H₃O⁺ (aq) Ka = 1.8 x 10⁻⁵

$\tt Ka=\dfrac{[C_2H_3O^{2-}[H_3O^+]]}{[HC_2H_3O_2]}}\\\\1.8\times 10^{-5}=\dfrac{0.22\times [H_3O^+]}{0.035}$

[H₃O⁺]=2.86 x 10⁻⁶ M

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If you combine 27.1 g of a solute that has a molar mass of 27.1 g/mol with 100.0 g of a solvent, what is the molality of the res

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<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

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3 years ago

Read 2 more answers

If the volume of a cylinder is reduced from 8.0 liters to 4.0 liters, the pressure of the gas in the cylinder will change from 7

Elza [17]

Answer:

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Given data:

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Final pressure = ?

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3 years ago

A 360. g iron rod is placed into 750.0 g of water at 22.5°C. The water temperature rises to 46.7°C. What was the initial tempera

Grace [21]

Answer: The initial temperature of the iron was $515^0C$

Explanation:

$heat_{absorbed}=heat_{released}$

As we know that,

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ .................(1)

where,

q = heat absorbed or released

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$m_2$ = mass of water = 750 g

$T_{final}$ = final temperature = $46.7^0C$

$T_1$ = temperature of iron = ?

$T_2$ = temperature of water = $22.5^oC$

$c_1$ = specific heat of iron = $0.450J/g^0C$

$c_2$ = specific heat of water= $4.184J/g^0C$

Now put all the given values in equation (1), we get

$-360\times 0.450\times (46.7-x)=[750\times 4.184\times (46.7-22.5)]$

$T_i=515^0C$

Therefore, the initial temperature of the iron was $515^0C$

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