Question

Question 4 (1 point) 0.444 g of an unknown diprotic acid is dissolved in about 60 mL of water in a beaker. The solution is transferred to a 100.00 mL volumetric flask, which is then filled up to the mark. The solution is mixed by inverting multiple times. 25.00 mL of this solution is then transferred to an Erlenmeyer flask for the titration. What mass of the diprotic acid is in the 25.00 mL that is transferred? Provide your answer to the correct number of significant figures. Your Answer: Answer units Question 5 (1 point) 0.377 g of an unknown triprotic acid are used to make a 100.00 mL solution. Then 25.00 mL of this solution is transferred to an Erlenmeyer flask and is titrated with 0.1197 M NaOH. The initial burette reading is 1.63 mL; when the titration endpoint is reached, the final burette reading is 35.41 mL. How many moles of triprotic acid are neutralized during the titration? Provide your answer in decimal form (e.g. 0.123) to the correct number of significant figures. Your Answer: Answer units

Answer 1

Mass of diprotic acid = \(0.444 \mathrm{~g}\) Volume \(=100.00 \mathrm{~mL}\)

Concentration of the solution initially \(=0.444 \mathrm{~g} / 100.00 \mathrm{~mL}=0.00444 \mathrm{~g} / \mathrm{mL}\)

Volume of the solution taken \(=25.00 \mathrm{~mL}\)

Mass of diprotic acid in \(25.00 \mathrm{~mL}==0.00444 \mathrm{~g} / \mathrm{mL} * 25.00 \mathrm{~mL}=0.111 \mathrm{~g}\)

since 0.444 has 3 sig.fig. and others have 4 sig.fig., the answer must be rounded to 3 sig.fig.

Mass of diprotic acid = \(0.111 \mathrm{~g}\)

The balanced equation is

$$ \mathrm{H}_{3} \mathrm{~A}+3 \mathrm{NaOH}-\mathrm{Na}_{3} \mathrm{~A}+3 \mathrm{H} 2 \mathrm{O} $$

Volume of \(\mathrm{NaOH}=35.41 \mathrm{~mL}-1.63 \mathrm{~mL}=33.78 \mathrm{~mL}\)

Molarity of \(\mathrm{NaOH}=0.1197 \mathrm{M}\)

Number of moles of \(\mathrm{NaOH}=0.1197 \mathrm{M} * 33.78 \mathrm{~mL}=4.043466 \mathrm{mmol}\)

According to balanced equation, 3 moles of \(\mathrm{NaOH}\) neutralizes 1 mole of \(\mathrm{H}_{3} \mathrm{~A}\) So, 4.043466 mmol of \(\mathrm{NaOH}\) would neutralize \(4.043466 \mathrm{mmol} / 3=1.347822 \mathrm{mmol}\) of \(\mathrm{H}_{3} \mathrm{~A}\)

Number of moles of \(\mathrm{H}_{3} \mathrm{~A}=1.347822 \mathrm{mmol}=\mathbf{0 . 0 0 1 3 5} \mathrm{mol}\)