Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

answer all questions. all your burette readings (initial and final) as …

Question

answer all questions. all your burette readings (initial and final) as well as the volume of your pipette must be recorded but no account of experimental procedure is required. all calculations must be done in this booklet. a is a solution containing 9.45 g dm⁻³ of impure ethanedioic acid. b is 0.150 mol dm⁻³ sodium hydroxide solution. (a) put a into the burette and titrate it against 20 or 25 cm³ portion of b using phenolphthalein as indicator. repeat the titration to obtain concordant titre values. tabulate your results and calculate the average volume of a used. (b) write a balanced chemical equation for the reaction. (c) from your results and the information provided, calculate the: (i) concentration of pure ethanedioic acid in a in mol dm⁻³; (ii) concentration of pure ethanedioic acid in a in g dm⁻³; (iii) percentage purity of the ethanedioic acid. (d) (i) what would be the colour of methyl orange indicator in solution b? (ii) give the reason why methyl orange is not a suitable indicator for the titration? h = 1.0, c = 12.0, o = 16.0, na = 23.0 25 marks

Explanation:

Step1: Write balanced equation

The reaction between ethanedioic acid ($H_2C_2O_4$) and sodium hydroxide ($NaOH$) is $H_2C_2O_4 + 2NaOH
ightarrow Na_2C_2O_4+2H_2O$.

Step2: For (c)(i) - Calculate moles of $NaOH$

Let's assume the volume of $B$ (sodium hydroxide) used is $V_{B}$ in $cm^3$ and its concentration $c_{B}=0.150\ mol\ dm^{-3}$. First convert volume to $dm^3$ ($V_{B}(dm^3)=\frac{V_{B}(cm^3)}{1000}$). Moles of $NaOH$, $n_{NaOH}=c_{B}\times V_{B}(dm^3)$. From the balanced - equation, the mole ratio of $H_2C_2O_4$ to $NaOH$ is $1:2$. So moles of $H_2C_2O_4$, $n_{H_2C_2O_4}=\frac{n_{NaOH}}{2}$. If the volume of $A$ (ethanedioic acid solution) used is $V_{A}$ in $dm^3$, then the concentration of pure ethanedioic acid in $A$, $c_{A}=\frac{n_{H_2C_2O_4}}{V_{A}}$.

Step3: For (c)(ii) - Calculate mass of pure $H_2C_2O_4$

The molar mass of $H_2C_2O_4$, $M = 2\times1.0+2\times12.0 + 4\times16.0=90.0\ g\ mol^{-1}$. Mass of pure $H_2C_2O_4$, $m = n_{H_2C_2O_4}\times M$. The concentration of pure ethanedioic acid in $A$ in $g\ dm^{-3}$ is $\frac{m}{V_{A}(dm^3)}$.

Step4: For (c)(iii) - Calculate percentage purity

Percentage purity $=\frac{\text{mass of pure acid}}{\text{mass of impure acid}}\times100\%$. Given the mass of impure acid is $9.35\ g\ dm^{-3}$.

Step5: For (d)(i) - Determine color of methyl orange in $B$

Sodium hydroxide is a strong base. Methyl orange has a color - change range of pH 3.1 - 4.4. In a basic solution (solution $B$), methyl orange is yellow.

Step6: For (d)(ii) - Explain unsuitability of methyl orange

Methyl orange changes color in an acidic pH range (3.1 - 4.4). The equivalence - point of the reaction between a weak acid (ethanedioic acid) and a strong base (sodium hydroxide) is in a basic pH range. So methyl orange will change color before the equivalence - point is reached, giving an inaccurate titration result.

Answer:

(a) Follow the titration procedure as described, record readings and calculate average volume of $A$ used.
(b) $H_2C_2O_4 + 2NaOH
ightarrow Na_2C_2O_4+2H_2O$
(c)(i) Depends on titration results (calculated as above).
(ii) Depends on titration results (calculated as above).
(iii) Depends on titration results (calculated as above).
(d)(i) Yellow
(ii) Methyl orange changes color in acidic pH range, while equivalence - point of weak acid - strong base titration is in basic pH range, leading to inaccurate results.