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Chapter 16: Problem 27

A 0.2688 -g sample of a monoprotic acid neutralizes \(16.4 \mathrm{~mL}\) of \(0.08133 \mathrm{M}\) KOH solution. Calculate the molar mass of the acid.

Short Answer

Expert verified
The molar mass of the acid is \(201.27 \mathrm{~g/mol}\)

Step by step solution

01

Convert milliliters of KOH to liters

First, convert the volume of the potassium hydroxide (KOH) solution from milliliters to liters because molarity is moles/liter. That is \(16.4 \mathrm{~mL} = 16.4 \times 10^{-3} \mathrm{~L} \)
02

Calculate the moles of KOH

Next, calculate the number of moles of KOH. Multiply the volume (in liters) of KOH by its molarity to get the moles of KOH. This results in \((16.4 \times 10^{-3}) \times 0.08133 = 1.335912 \times 10^{-3}\) moles of KOH.
03

Identify the moles of the unknown acid

In a neutralization reaction, a monoprotic acid reacts with KOH in a 1:1 ratio. That is, every mole of KOH neutralizes one mole of the anonymous acid. So the moles of the anonymous acid is equal to the moles of KOH, which is \(1.335912 \times 10^{-3}\) moles.
04

Calculate the molar mass of the acid

Finally, to get the molar mass, divide the mass of the acid by the moles of the acid. This gives \[ \frac{0.2688}{1.335912 \times 10^{-3}} = 201.27 \mathrm{~g/mol}\]

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Most popular questions from this chapter

\(\mathrm{CaSO}_{4}\left(K_{\mathrm{sp}}=2.4 \times 10^{-5}\right)\) has a larger \(K_{\mathrm{sp}}\) value than that of \(\mathrm{Ag}_{2} \mathrm{SO}_{4}\left(K_{\mathrm{sp}}=1.4 \times 10^{-5}\right) .\) Does it follow that \(\mathrm{CaSO}_{4}\) also has greater solubility \((\mathrm{g} / \mathrm{L}) ?\)

Acid-base reactions usually go to completion. Confirm this statement by calculating the equilibrium constant for each of the following cases: (a) a strong acid reacting with a strong base; (b) a strong acid reacting with a weak base \(\left(\mathrm{NH}_{3}\right) ;\) (c) a weak acid (CH \(_{3} \mathrm{COOH}\) ) reacting with a strong base; (d) a weak acid (CH \(_{3}\) COOH) reacting with a weak base \(\left(\mathrm{NH}_{3}\right) .\) (Hint: Strong acids exist as \(\mathrm{H}^{+}\) ions and strong bases exist as \(\mathrm{OH}^{-}\) ions in solution. You need to look up \(K_{a}, K_{b},\) and \(K_{w^{\prime}}\) )

A 0.1276 -g sample of an unknown monoprotic acid was dissolved in \(25.0 \mathrm{~mL}\) of water and titrated with \(0.0633 \mathrm{M} \mathrm{NaOH}\) solution. The volume of base required to bring the solution to the equivalence point was \(18.4 \mathrm{~mL}\). (a) Calculate the molar mass of the acid. (b) After \(10.0 \mathrm{~mL}\) of base had been added during the titration, the \(\mathrm{pH}\) was determined to be 5.87 . What is the \(K_{\mathrm{a}}\) of the unknown acid?

How can we predict whether a precipitate will form when two solutions are mixed?

A diprotic acid, \(\mathrm{H}_{2} \mathrm{~A},\) has the following ionization constants: \(K_{a_{1}}=1.1 \times 10^{-3}\) and \(K_{\mathrm{a}_{2}}=2.5 \times 10^{-6} . \mathrm{In}\) order to make up a buffer solution of \(\mathrm{pH} 5.80\), which combination would you choose: NaHA/ \(\mathrm{H}_{2} \mathrm{~A}\) or \(\mathrm{Na}_{2} \mathrm{~A} / \mathrm{NaHA} ?\)
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