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miskamm [114]
1 year ago
12

An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4. What is the concentration of hydronium ion in

this solution (m)?
Chemistry
1 answer:
Alex_Xolod [135]1 year ago
5 0

An aqueous basic solution has a concentration of 0. 050 m and kb is 4. 4 × 10^-4, the concentration of hydronium ion in this solution (m) is 2.234 × 10⁻¹² M.

Methylamine is an amine which is an organic weak base. Its chemical formula is CH₃NH₂. When it undergoes hydrolysis wherein water is acting as an acid, the reaction would be

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Then, we use the ICE analysis which stands for Initial-Change-Equilibrium.

CH₃NH₂ + H₂O ⇆  CH₃NH₃ + OH⁻

Initial                    0.05          -                 0         0

Change                 -x                              +x       +x

----------------------------------------------------------------------------

Equilibrium         0.05-x                           x          x

Then, we use the equation for the equilibrium constant of basicity.

Kb = [CH₃NH₃][OH⁻]/[CH₃NH₂] = 4.4×10⁻⁴

4.4×10⁻⁴ = [x][x]/[0.05-x]

[x] = 4.4756×10⁻³

Here, variable x denotes the number of moles of the substance which is involved in the reaction. Since the equilibrium amount of OH⁻ is equal to x, then the concentration of OH⁻ is also 4.4756×10⁻³. Thus,

pOH = -log[OH⁻]

pOH = -log[4.4756×10⁻³] = 2.35

The relationship between pOH and pH is that pH + pOH = 14. Thus,

pH = 14 - 2.35 = 11.65

pH = -log[H⁺]

11.65 = -log[H⁺]

[H⁺] = 2.234 × 10⁻¹² M

Thus, we find the concentration of solution is 2.234 × 10⁻¹² M.

Learn more about aqueous solution: brainly.com/question/11097800

#SPJ4

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a. Molecular equation of the reaction is given below :

CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol

moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaCl₂.2 H₂O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of K₂C₂O₄.H₂O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

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mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g

mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%

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