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Saccharin, a sugar substitute, is a weak acid with \(\mathrm{p} K_{a}=2.32\) at \(25^{\circ} \mathrm{C}\). It ionizes in aqueous solution as follows: \(\mathrm{HNC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{NC}_{7} \mathrm{H}_{4} \mathrm{SO}_{3}^{-}(a q)\) What is the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of this substance?

Short Answer

Expert verified
The pH of the 0.10 M saccharin solution can be calculated using the following steps: 1. Calculate the Ka from the given pKa value: \(Ka = 10^{-2.32}\) 2. Set up the ICE table and write the expression for Ka. 3. Solve the equation for x, assuming x is much smaller than 0.10: \(x^2 = Ka(0.10)\) 4. Calculate the equilibrium concentration of H鈦 ions (x): \(x =\sqrt{(10^{-2.32})(0.10)}\) 5. Calculate the pH of the solution using the calculated x value: \(pH = -\log{x}\)

Step by step solution

01

Find the Ka from the given pKa value

Using the relationship between Ka and pKa, where pKa = -log(Ka), we calculate the Ka for saccharin: \[Ka = 10^{-pKa} = 10^{-2.32}\]
02

Write the balanced ionization reaction and set up the ICE table

The ionization reaction is given as: \[HNC_7H_4SO_3(aq) \rightleftharpoons H^+(aq) + NC_7H_4SO_3^鈭(aq)\] The ICE table can be set up as: | | Initial (M) | Change (M) | Equilibrium (M) | |-------------------|-------------|------------|-----------------| | HNC鈧嘓鈧凷O鈧 | 0.10 | -x | 0.10 - x | | H鈦 | 0 | +x | x | | NC鈧嘓鈧凷O鈧冣伝 | 0 | +x | x | We want to find the concentration of H鈦 ions at equilibrium, which is x.
03

Write the expression for Ka and substitute the equilibrium concentrations

Ka is the acid-dissociation constant: \[Ka = \frac{[H^+][NC_{7}H_{4}SO_{3}^{-}]}{[HNC_{7}H_{4}SO_{3}]}\] Substitute the equilibrium concentrations from the ICE table: \[Ka = \frac{x^2}{0.10 - x}\] Solve this equation for x: \[x^2 = Ka(0.10 - x)\] Because saccharin is a weak acid, we can assume that x is much smaller than 0.10, so we have: \[x^2 = Ka(0.10)\]
04

Calculate the equilibrium concentration of H鈦 ions (x)

Now we can solve for x using the Ka value calculated in Step 1: \[x =\sqrt{Ka(0.10)} = \sqrt{(10^{-2.32})(0.10)}\]
05

Calculate the pH of the solution

Use the concentration of H鈦 ions to calculate the pH: \[pH = -\log[H^+] = -\log{x}\] After plugging in the value of x from Step 4, we can find the pH of the 0.10 M saccharin solution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Weak Acid Ionization
Understanding the behavior of weak acids in water is essential when working with pH calculations. Unlike strong acids which fully dissociate, weak acids only partially ionize in solution. This partial ionization is what makes the acid 'weak' and it鈥檚 represented by a reversible reaction. For example, saccharin (a weak acid) ionizes in water to form hydrogen ions (H鈦) and the saccharinate anion (NC鈧嘓鈧凷O鈧冣伝).

The degree of ionization is not only determined by the acid's inherent strength but also influenced by the concentration of the acid and the properties of the solvent. In dilute solutions, weak acids are less dissociated, as evidenced by the equilibrium that sets up between the non-ionized acid and the ions produced.
Ka and pKa Relationship
The strength of a weak acid is quantified by its acid-dissociation constant (Ka). The larger the Ka, the stronger the acid, meaning it ionizes more in solution. However, working with these constant values when they are extremely small can be cumbersome. To simplify calculations, chemists often use the pKa value, which is the negative base-10 logarithm of the Ka value: \( pKa = -\log(Ka) \).

This relationship also works the other way around, allowing you to find Ka if you know the pKa, which is particularly useful in pH calculations. The pKa provides a more convenient scale for expressing the strength of an acid. In the case of saccharin with a pKa of 2.32, we can calculate its Ka using \( Ka = 10^{-pKa} \) to find the concentration of hydrogen ions at equilibrium.
ICE Table
An ICE table, standing for Initial, Change, and Equilibrium, is an organizational tool used to keep track of the changes in concentrations of reactants and products as a chemical reaction approaches equilibrium. For weak acid ionizations, we start by listing the initial concentrations of the acid and ions, the changes that will occur during reaction, and the concentrations at equilibrium.

When dealing with weak acid ionization, it is generally safe to assume that the change in concentration of the acid (\(x\)) is small compared to the initial concentration, leading to simplifications in calculations. The ICE table helps in clearly defining the relationships between the concentrations of different species involved in the reaction and is instrumental in setting up the equation to calculate Ka.
Acid-Dissociation Constant
The acid-dissociation constant (Ka) provides a measure of the extent to which an acid can donate protons in an aqueous solution. The formula for Ka is \( Ka = \frac{[H^+][A^-]}{[HA]} \), where [H鈦篯 is the concentration of hydrogen ions, [A鈦籡 is the concentration of the acid's anion, and [HA] is the concentration of the acid itself.

For weak acids like saccharin, Ka is relatively small, indicating limited ionization. By substituting the equilibrium concentrations derived from the ICE table into the Ka expression, we can solve for the concentration of hydrogen ions. This, in turn, enables us to calculate the pH of the solution. The understanding of Ka is pivotal in making these types of calculations and understanding acid-base equilibria.

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(0.0167 \mathrm{M} \mathrm{HNO}_{3},\) (b) \(0.225 \mathrm{~g}\) of \(\mathrm{HClO}_{3}\) in \(2.00 \mathrm{~L}\) of solution, (c) \(15.00 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\) diluted to \(0.500 \mathrm{~L}, (\mathrm{~d})\) a mixture formed by adding \(50.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{HCl}\) to \(125 \mathrm{~mL}\) of \(0.010 \mathrm{M} \mathrm{HI}\)

Carbon dioxide in the atmosphere dissolves in raindrops to produce carbonic acid \(\left(\mathrm{H}_{2} \mathrm{CO}_{3}\right),\) causing the \(\mathrm{pH}\) of clean, \(\mathrm{un}^{-}\) polluted rain to range from about 5.2 to \(5.6 .\) What are the ranges of \(\left[\mathrm{H}^{+}\right]\) and \(\left[\mathrm{OH}^{-}\right]\) in the raindrops?

(a) The hydrogen oxalate ion \(\left(\mathrm{HC}_{2} \mathrm{O}_{4}^{-}\right)\) is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conjugate acid of \(\mathrm{HC}_{2} \mathrm{O}_{4}\) ? What is its conjugate base?

Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) In general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) In a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid \(\left(\mathrm{H}_{2} \mathrm{Te}\right)\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{~S}\) because Te is more electronegative than \(\mathrm{S}\).

Calculate \(\left[\mathrm{H}^{+}\right]\) for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) \(\left[\mathrm{OH}^{-}\right]=0.00045 \mathrm{M} ;\) (b) \(\left[\mathrm{OH}^{-}\right]=8.8 \times 10^{-9} \mathrm{M} ;(\mathrm{c})\) a so- lution in which \(\left[\mathrm{OH}^{-}\right]\) is 100 times greater than \(\left[\mathrm{H}^{+}\right]\).

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