What is the acetate ion concentration, [C2H3O2 1-] of a 100mL of calcium acetate solution, Ca(C2H3O2)2, with 25.0g dissolved Ca(C2H3O2)2?
What is [H?] in a 0.270 M solution of acrylic acid, CH2CHCOOH (Ka = 3.16 × 10?5)?
The Correct Answer and Explanation is :
Question 1: Acetate Ion Concentration
Given Data:
- Calcium acetate: ( \text{Ca(C}_2\text{H}_3\text{O}_2)_2 )
- Mass of calcium acetate = ( 25.0 \, \text{g} )
- Volume of solution = ( 100.0 \, \text{mL} = 0.100 \, \text{L} )
Steps:
- Molar Mass of Calcium Acetate:
( \text{Ca(C}_2\text{H}_3\text{O}_2)_2 ):
( 40.08 + 2(12.01 + 3(1.01) + 2(16.00)) = 158.17 \, \text{g/mol} ). - Moles of Calcium Acetate:
[
\text{moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{25.0}{158.17} \approx 0.158 \, \text{mol}.
] - Concentration of Calcium Acetate:
[
[\text{Ca(C}_2\text{H}_3\text{O}_2)_2] = \frac{\text{moles}}{\text{volume}} = \frac{0.158}{0.100} = 1.58 \, \text{M}.
] - Acetate Ion Concentration:
Each calcium acetate dissociates into 2 acetate ions (( \text{C}_2\text{H}_3\text{O}_2^- )).
[
[\text{C}_2\text{H}_3\text{O}_2^-] = 2 \times 1.58 = 3.16 \, \text{M}.
]
Answer: ( [\text{C}_2\text{H}_3\text{O}_2^-] = 3.16 \, \text{M} ).
Question 2: [H⁺] in Acrylic Acid Solution
Given Data:
- ( \text{CH}_2\text{CHCOOH} ), ( K_a = 3.16 \times 10^{-5} )
- Initial concentration (( C_0 )) = ( 0.270 \, \text{M} )
Steps:
- Expression for ( K_a ):
[
K_a = \frac{[H^+][A^-]}{[HA]}
]
Assuming ( [H^+] = [A^-] = x ) and ( [HA] = C_0 – x ):
[
K_a = \frac{x^2}{C_0 – x}.
] - Simplification (since ( K_a ) is small):
( C_0 – x \approx C_0 ).
[
K_a \approx \frac{x^2}{C_0}.
] - Solve for ( x ):
[
x = \sqrt{K_a \cdot C_0}.
]
Substituting values:
[
x = \sqrt{(3.16 \times 10^{-5})(0.270)} \approx \sqrt{8.532 \times 10^{-6}} \approx 2.92 \times 10^{-3}.
]
Answer:
( [H^+] = 2.92 \times 10^{-3} \, \text{M} ).
Explanation (300 Words)
For the acetate ion concentration problem, calcium acetate (( \text{Ca(C}_2\text{H}_3\text{O}_2)_2 )) dissociates completely in water into one calcium ion (( \text{Ca}^{2+} )) and two acetate ions (( \text{C}_2\text{H}_3\text{O}_2^- )). The total concentration of acetate ions is twice the molarity of calcium acetate. Calculations require converting the given mass of the salt into moles using its molar mass, then dividing by the solution volume. Since the dissociation is complete, the acetate ion concentration equals ( 2 \times \text{[Ca(C}_2\text{H}_3\text{O}_2)_2] ).
For the acrylic acid (( \text{CH}_2\text{CHCOOH} )) problem, we use the weak acid dissociation equilibrium constant (( K_a )) to determine the concentration of hydrogen ions. Weak acids partially dissociate in water, and ( K_a ) quantifies this dissociation. Assuming ( x ) represents the concentration of ( H^+ ), the equilibrium expression simplifies to ( K_a = x^2 / C_0 ). Given the small ( K_a ), the approximation ( C_0 – x \approx C_0 ) avoids quadratic equations. Calculating ( x ), we find ( [H^+] = 2.92 \times 10^{-3} \, \text{M} ). This method highlights the relationship between ( K_a ), initial acid concentration, and ( [H^+] ).