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3 years ago

Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5

Chemistry

1 answer:

lubasha [3.4K]3 years ago

4 0

The pH of the solution is 2.54.

Explanation:

pH is the measure of acidity of the solution and Ka is the dissociation constant. Dissociation constant is the measure of concentration of hydrogen ion donated to the solution.

The solution of C₆H₂O₆ will get dissociated as C₆HO₆ and H+ ions. So the molar concentration of 0.1 M is present at the initial stage. Lets consider that the concentration of hydrogen ion released as x and the same amount of the base ion will also be released.

So the dissociation constant Kₐ can be written as the ratio of concentration of products to the concentration of reactants. As the concentration of reactants is given as 0.1 M and the concentration of products is considered as x for both hydrogen and base ion. Then the

$K_{a}=\frac{[H^{+}][HB] }{[reactant]}$

[HB] is the concentration of base.

$8 * 10^{-5} =\frac{x^{2} }{0.1}\\\\\\x^{2} = 8 * 10^{-5}*0.1$

$x^{2} = 0.08 * 10^{-4}\\ \\x = 0.283*10^{-2}$

Then

$pH = - log [x] = - log [ 0.283 * 10^{-2}]\\ \\pH = 2 + 0.548 = 2.54$

So the pH of the solution is 2.54.

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Suppose a 0.025M aqueous solution of sulfuric acid (H2SO4) is prepared. Calculate the equilibrium molarity of SO4−2. You'll find

FromTheMoon [43]

Answer: The concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

Explanation:

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

$H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)$

0.025 0.025 0.025

Equation for the second dissociation of sulfuric acid:

$HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)$

Initial: 0.025 0.025

At eqllm: 0.025-x 0.025+x x

The expression of second equilibrium constant equation follows:

$Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}$

We know that:

$Ka_2\text{ for }H_2SO_4=0.01$

Putting values in above equation, we get:

$0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608$

Neglecting the negative value of 'x', because concentration cannot be negative.

So, equilibrium concentration of sulfate ion = x = 0.00608 M

Hence, the concentration of $SO_4^{2-}$ at equilibrium is 0.00608 M

4 0

3 years ago

How many milliliters of 0.200 m fecl3 are needed to react with an excess of na2s to produce 1.38 g of fe2s3 if the percent yield

ahrayia [7]

Answer: 100 ml


Explanation:

1) Convert 1.38 g of Fe₂S₃ into number of moles, n

i) Formula: n = mass in grass / molar mass

ii) molar mass of Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol


iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of Fe₂S₃


2) Use the percent yield to calculate the theoretical amount:


65% = 0.65 = actual yield/ theoretical yield =>

theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol Fe₂S₃

3) Chemical equation:


 3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

4) Stoichiometrical mole ratios:


3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl

5) Proportionality:

2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃

=> x = 0.020 mol FeCl₃

6) convert 0.020 mol to volume


i) Molarity formula: M = n / V


ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml

3 0

3 years ago

Read 2 more answers

Calcule la densidad del hidrógeno H2 en g/L a 327 mm Hg y 48ºc

tankabanditka [31]

<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm , N/m²

V = volume, liter

n = number of moles

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin

n = N / No

n = mole

No = Avogadro number (6.02.10²³)

n = m / MW

m = mass

MW = molecular weight

For density , can be formulated :

$\tt \rho=\dfrac{P\times MW}{R\times T}$

P = 327 mmHg = 0,430263 atm

R = 0.082 L.atm / mol K

T = 48 ºC = 321.15 K

MW of H₂ = 2.015 g/mol

The density :

$\rho=\dfrac{0,430263\times 2.015 }{0.082\times 321.15}\\\\\rho=0.033~g/L$

4 0

2 years ago

5. 16.3 g of NaCl is dissolved in water to make 1.75 L of solution. What is the molarity of this solution? A 0.159 M B 0.278 M C

ira [324]

Answer: The molarity of this solution is 0.159 M.

Explanation:

Given: Mass of solute = 16.3 g

Volume = 1.75 L

Number of moles is defined as the mass of substance divided by its molar mass.

Hence, moles of NaCl (molar mass = 58.44 g/mol) ar calculated as follows.

$Moles = \frac{mass}{molar mass}\\= \frac{16.3 g}{58.44 g/mol}\\= 0.278 mol$

Molarity is the number of moles of a substance present in a liter of solution.

So, molarity of the given solution is calculated as follows.

$Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.278 mol}{1.75}\\= 0.159 M$

Thus, we can conclude that the molarity of this solution is 0.159 M.

5 0

3 years ago

What are three limitations of current cloaking technology.

kirill [66]

Explanation:

1.undetectable to electromagnetic waves

2.hiding an object from an illumination containing diffre t wave lengths become difficult as the object sizes grow.

3. reduce the scattering by two orders.

4 0

3 years ago

Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5​

Find the pH of a 0.100 molar H2C6O6 solution with ka, where KA is equal 8.0×10–5