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Chapter 17: Problem 90

A sample of 0.2140 \(\mathrm{g}\) of an unknown monoprotic acid was dissolved in 25.0 \(\mathrm{mL}\) of water and titrated with 0.0950 \(\mathrm{M}$$ \mathrm{NaOH}\). The acid required 30.0 \(\mathrm{mL}\) of base to reach the equivalence point. (a) What is the molar mass of the acid? (b) After 15.0 \(\mathrm{mL}\) of base had been added in the titration, the pH was found to be \(6.50 .\) What is the \(K_{a}\) for the unknown acid?

Short Answer

Expert verified
The molar mass of the unknown acid is calculated as follows: \(molar\ mass = \frac{0.2140 g}{(0.0950 M \times 0.0300 L)}\). After finding the moles of the unknown acid, we determine the Ka using the formula \(Ka = 10^{-6.50}\).

Step by step solution

01

Calculate moles of NaOH used to reach the equivalence point

At the equivalence point, the moles of the unknown acid equal the moles of NaOH. We are given the volume and concentration of NaOH: Volume of NaOH = 30.0 mL (convert to L by dividing by 1000) Concentration of NaOH = 0.0950 M We can find the moles of NaOH using the formula: moles = concentration × volume: moles of NaOH = 0.0950 M × 0.0300 L
02

Step 2:Find the moles of the unknown acid

Since the moles of the unknown acid equal the moles of NaOH at the equivalence point, we can say: moles of unknown acid = moles of NaOH
03

Calculate the molar mass of the unknown acid

To find the molar mass of the acid, we will use the formula: molar mass = mass / moles We are given the mass of the unknown acid (0.2140g), and we have just calculated the moles. Now we can solve for molar mass: molar mass = 0.2140 g / moles of unknown acid
04

Use the titration halfway point to find the Ka of the acid

At the halfway point of the titration, the pH equals the pKa of the acid. In this case, the pH was found to be 6.50, so pKa = 6.50. Ka is the antilogarithm of the pKa, so to find the Ka, use the formula: Ka = 10^(-pKa) Ka = 10^(-6.50)

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Most popular questions from this chapter

Rainwater is acidic because \(\mathrm{CO}_{2}(\mathrm{g})\) dissolves in the water, creating carbonic acid, \(\mathrm{H}_{2} \mathrm{CO}_{3}\) . If the rainwater is too acidic, it will react with limestone and seashells (which are principally made of calcium carbonate, CaCO_ \(_{3} ) .\) Calculate the concentrations of carbonic acid, bicarbonate ion \(\left(\mathrm{HCO}_{3}^{-}\right)\) and carbonate ion \(\left(\mathrm{CO}_{3}^{2-}\right)\) that are in a raindrop that has a pH of 5.60 , assuming that the sum of all three species in the raindrop is \(1.0 \times 10^{-5} M .\)

Tooth enamel is composed of hydroxyapatite, whose simplest formula is \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) , and whose corresponding \(K_{\mathrm{sp}}=6.8 \times 10^{-27}\) . As discussed in the Chemistry and Life box on page \(746,\) fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{F},\) whose \(K_{s p}=1.0 \times 10^{-60}\) (a) Write the expression for the solubility-constant for hydroxyapatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds.

Mathematically prove that the \(\mathrm{pH}\) at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equivalence point) is equal to p \(K_{a}\) for the acid.

Consider a beaker containing a saturated solution of \(\mathrm{CaF}_{2}\) in equilibrium with undissolved \(\mathrm{CaF}_{2}(s)\). Solid \(\mathrm{CaCl}_{2}\) is then added to the solution. (a) Will the amount of solid \(\mathrm{CaF}_{2}\) at the bottom of the beaker increase, decrease, or remain the same? (b) Will the concentration of \(\mathrm{Ca}^{2+}\) ions in solution increase or decrease? (c) Will the concentration of \(\mathbf{F}^{-}\) ions in solution increase or decrease?

Baking soda (sodium bicarbonate, \(\mathrm{NaHCO}_{3}\)) reacts with acids in foods to form carbonic acid (\(\mathrm{H}_{2} \mathrm{CO}_{3}\)), which in turn decomposes to water and carbon dioxide gas. In a cake batter, the \(\mathrm{CO}_{2}(g)\) forms bubbles and causes the cake to rise. (a) A rule of thumb in baking is that 1\(/ 2\) teaspoon of baking soda is neutralized by one cup of sour milk. The acid component in sour milk is lactic acid, \(\mathrm{CH}_{3} \mathrm{CH (\mathrm{OH}) \)\mathrm{COOH}\( .Write the chemical equation for this neutralization reaction. (b) The density of baking soda is 2.16 \)\mathrm{g} / \mathrm{cm}^{3} .\( Calculate the concentration of lactic acid in one cup of sour milk(assuming the rule of thumb applies), in units of mol/L. (One cup \)=236.6 \mathrm{mL}=48\( teaspoons). (c) If 1/2 teaspoon of baking soda is indeed completely neutralized by the lactic acid in sour milk, calculate the volume of carbon dioxide gas that would be produced at 1 atm pressure, in an oven set to \)350^{\circ} \mathrm{F}$ .
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