<span>Answer: 0.094%
</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />
<span>Only the ionization of the formic acid is the important part.
</span><span />
<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />
<span>2) Mass balance:
</span><span />
<span> HCOOH(aq) HCOO⁻(aq) H⁺(aq).
Start 0.311 0.189
Reaction - x +x +x
Final 0.311 - x 0.189 + x x
3) Acid constant equation:
</span><span />
<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />
<span>= (0.189 + x )x / (0.311 - x) = 0.000177
4) Solve the equation:
You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />
<span>With that approximation the equation to solve becomes:
</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M
5) With that number, the percent of ionization (alfa) is:
</span><span />
<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
<span></span><span />
Answer:
0.404M
Explanation:
...<em>To make exactly 100.0mL of solution...</em>
Molar concentration is defined as the amount of moles of a solute (In this case, nitrate ion, NO₃⁻) in 1 L of solution.
To solve this question we need to convert the mass of Fe(NO₃)₃ to moles. As 1 mole of Fe(NO₃)₃ contains 3 moles of nitrate ion we can find moles of nitrate ion in 100.0mL of solution, and we can solve the amount of moles per liter:
<em>Moles Fe(NO₃)₃ -Molar mass: 241.86g/mol-:</em>
3.26g * (1mol / 241.86g) =
0.01348 moles Fe(NO₃)₃ * (3 moles of NO₃⁻ / 1mole Fe(NO₃)₃) =
<em>0.0404 moles of NO₃⁻</em>
In 100mL = 0.1L, the molar concentration is:
0.0404 moles of NO₃⁻ / 0.100L =
<h3>0.404M</h3>
