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devlian [24]
1 year ago
15

Determine the volume in mL of 0.242 M NaOH(aq) needed to reach the equivalence (stoichiometric) point in the titration of 46.79

mL of 0.204 M propanoic acid(aq). The Ka of propanoic acid is 1.3 x 10-5.
Chemistry
1 answer:
Alla [95]1 year ago
5 0

The volume (in mL) of 0.242 M NaOH solution needed for the titration reaction is 39.44 mL

<h3>Balanced equation </h3>

CH₃CH₂COOH + NaOH —> CH₃CH₂COONa + H₂O

From the balanced equation above,

  • The mole ratio of the acid, CH₃CH₂COOH (nA) = 1
  • The mole ratio of the base, NaOH (nB) = 1

<h3>How to determine the volume of NaOH</h3>
  • Volume of acid, CH₃CH₂COOH (Va) = 46.79 mL
  • Molarity of acid, CH₃CH₂COOH (Ma) = 0.204 M
  • Molarity of base, NaOH (Mb) = 0.242 M
  • Volume of base, KOH (Vb) =?

MaVa / MbVb = nA / nB

(0.204 × 46.79) / (0.242 × Vb) = 1

Cross multiply

0.242 × Vb = 0.204 × 46.79

Divide both side by 0.242

Vb = (0.204 × 46.79) / 0.242

Vb = 39.44 mL

Thus, the volume of NaOH needed for the reaction is 39.44 mL

Learn more about titration:

brainly.com/question/14356286

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Calculate the percent ionization of formic acid (hco2h) in a solution that is 0.311 m in formic acid and 0.189 m in sodium forma
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<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>
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