How many milliliters of 2.155 M KOH are required to titrate \(25.00 \mathrm{mL}\) of \(0.3057 \mathrm{M} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}\) (prop-ionic acid)?

Short Answer

Expert verified
3.544 milliliters of 2.155 M KOH are required to titrate 25 ml of 0.3057 M CH3CH2COOH (prop-ionic acid).

Step by step solution

01

Identify the Reaction

For every mole of CH3CH2COOH that reacts with KOH, it produces one mole of salt and one mole of water. So, the titration reaction is: CH3CH2COOH + KOH → CH3CH2COOK + H2O. According to this balanced reaction, the stoichiometric ratio of CH3CH2COOH to KOH is 1:1.
02

Calculate moles of CH3CH2COOH

We know the volume of CH3CH2COOH to be 25.00 ml or 0.025 liters and its molarity as 0.3057 M. Molarity is defined as the number of moles of solute per liter of solution, so we multiply the volume by the molarity to find the number of moles of CH3CH2COOH, which is \(0.025 L * 0.3057 M = 0.0076425 moles\)
03

Calculate the volume of KOH

Using the stoichiometry of the reaction (1:1), we know that we need an equal number of moles of KOH to titrate CH3CH2COOH. Given that the molarity of KOH is 2.155 M, we can find the required volume using the following formula: volume = moles / molarity. Therefore, the volume of KOH required in liters is \(0.0076425 moles / 2.155 M = 0.003544 liters\) or 3.544 milliliters.

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Most popular questions from this chapter

Determine the concentration of the ion indicated in each solution. (a) \(\left[\mathrm{K}^{+}\right]\) in \(0.238 \mathrm{M} \mathrm{KNO}_{3} ;\) (b) \(\left[\mathrm{NO}_{3}\right]\) in \(0.167 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} ;(\mathrm{c})\left[\mathrm{Al}^{3+}\right]\) in \(0.083 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3};\) (d) \(\left[\mathrm{Na}^{+}\right]\) in \(0.209 \mathrm{M} \mathrm{Na}_{3} \mathrm{PO}_{4}\).

Which solutions would you use to precipitate \(\mathrm{Mg}^{2+}\) from an aqueous solution of \(\mathrm{MgCl}_{2} ?\) Explain your choice. (a) \(\mathrm{KNO}_{3}(\mathrm{aq}) ;\) (b) \(\mathrm{NH}_{3}(\mathrm{aq}) ;\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq});\) (d) \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}(\mathrm{aq})\).

Which solution has the greatest \(\left[\mathrm{SO}_{4}^{2-}\right]:\) (a) \(0.075 \mathrm{M}\) \(\mathrm{H}_{2} \mathrm{SO}_{4} ; \quad\) (b) \(\quad 0.22 \mathrm{M} \mathrm{MgSO}_{4} ; \quad\) (c) \(\quad 0.15 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.080 \mathrm{M} \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} ;\) (e) \(0.20 \mathrm{M} \mathrm{CuSO}_{4} ?\)

The active ingredients in a particular antacid tablet are aluminum hydroxide, \(\mathrm{Al}(\mathrm{OH})_{3},\) and magnesium hydroxide, \(\mathrm{Mg}(\mathrm{OH})_{2} . \quad \mathrm{A} 5.00 \times 10^{2} \mathrm{mg}\) sample of the active ingredients was dissolved in \(50.0 \mathrm{mL}\) of \(0.500 \mathrm{M} \mathrm{HCl} .\) The resulting solution, which was still acidic, required \(16.5 \mathrm{mL}\) of \(0.377 \mathrm{M} \mathrm{NaOH}\) for neutralization. What are the mass percentages of \(\mathrm{Al}(\mathrm{OH})_{3}\) and \(\mathrm{Mg}(\mathrm{OH})_{2}\) in the sample?

When aqueous sodium carbonate, \(\mathrm{Na}_{2} \mathrm{CO}_{3}\), is treated with dilute hydrochloric acid, HCl, the products are sodium chloride, water, and carbon dioxide gas. What is the net ionic equation for this reaction?

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