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Chapter 15: Problem 17

Calculate the pH of each of the following solutions. a. \(0.100 M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) b. \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\) c. pure \(\mathrm{H}_{2} \mathrm{O}\) d. a mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 \mathrm{M}\) \(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)

Short Answer

Expert verified
a. pH of \(0.1M\) propanoic acid: 2.96 b. pH of \(0.1M\) sodium propanoate: 9.44 c. pH of pure water: 7.00 d. pH of the mixture containing \(0.1M\) propanoic acid and \(0.1M\) sodium propanoate: 4.89

Step by step solution

01

Write the equilibrium expression and Ka

The equilibrium expression for the dissociation of propanoic acid is: \(\mathrm{HC_3H_5O_2} \rightleftharpoons \mathrm{H^+} + \mathrm{C_3H_5O_2^-}\) Ka = \(\frac{[\mathrm{H^+}][\mathrm{C_3H_5O_2^-}]}{[\mathrm{HC_3H_5O_2}]}\)
02

Setup the ice table and solve for [H+]

Create an equilibrium ice table: Initial: \(0.100M\) \(0\) \(0\) Change: \(-x\) \(+x\) \(+x\) Equilibrium: \(.1-x\) \(x\) \(x\) Now substitute equilibrium values into the Ka expression: \(1.3 \times 10^{-5} = \frac{x^2}{0.1-x}\) Solving the quadratic equation, we get x = $[\mathrm{H^+}] = 1.1 \times 10^{-3}\)
03

Calculate the pH

Finally, we use the pH formula: pH = -log([H+]) pH = -log(\(1.1 \times 10^{-3}\)) ≈ 2.96 b. Calculate the pH of \(0.100 M\) sodium propanoate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\right)\)
04

Determine the conjugate acid- base pair and solve for Ka

Sodium propanoate's formula is \(\mathrm{NaC_3H_5O_2}\). It is the salt formed from the weak acid \(\mathrm{HC_3H_5O_2}\) we dealt with in the previous solution. The latter has a Ka value of \(1.3 \times 10^{-5}\).
05

Calculate the Kb value

Using the relationship between Ka and Kb for a conjugate acid-base pair: Ka × Kb = Kw. Solve for Kb: Kb = \(\frac{K_w}{K_a}\) = \(\frac{1.0 \times 10^{-14}}{1.3 \times 10^{-5}}\) ≈ 7.69 × 10^(-10)
06

Write the equilibrium expression and setup an ICE table

The equilibrium expression for the dissociation of propanoate ion is: \(\mathrm{C_3H_5O_2^-} + \mathrm{H_2O} \rightleftharpoons \mathrm{OH^-} + \mathrm{HC_3H_5O_2}\) Kb = \(\frac{[\mathrm{OH^-}][\mathrm{HC_3H_5O_2}]}{[\mathrm{C_3H_5O_2^-}]}\) Create an equilibrium ice table: Initial: \(0.100M\) \(0\) \(0\) Change: \(-x\) \(+x\) \(+x\) Equilibrium: \(.1-x\) \(x\) \(x\) Now substitute equilibrium values into the Kb expression: \(7.69 \times 10^{-10} = \frac{x^2}{0.1-x}\) Solving the quadratic equation, we get x = $[\mathrm{OH^-}] = 2.77 \times 10^{-5}\)
07

Calculate the pOH and then pH

pOH = -log(\(2.77 \times 10^{-5}\)) ≈ 4.56 pH = 14 - pOH ≈ 14 - 4.56 ≈ 9.44 c. Calculate the pH of pure \(\mathrm{H}_{2} \mathrm{O}\)
08

Apply the pH formula for pure water

Since pure water is neutral, we know that its pH is 7. d. Calculate the pH of a mixture containing \(0.100 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(0.100 \mathrm{M} \mathrm{NaC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}\)
09

Use Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation for a buffer solution is: pH = pKa + log(\(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\)) The initial concentrations of \(\mathrm{HC_3H_5O_2}\) and \(\mathrm{NaC_3H_5O_2}\) (which will dissociate into \(\mathrm{C_3H_5O_2^-}\)) are equal.
10

Calculate the pH

pH = pKa + log(\(\frac{[C_3H_5O_2^-]}{[HC_3H_5O_2]}\)) = -log(\(1.3 \times 10^{-5}\)) + log(1) = 4.89 Summary: a. pH of .1M propanoic acid: 2.96 b. pH of .1M sodium propanoate: 9.44 c. pH of pure water: 7.00 d. pH of the mixture containing .1M propanoic acid and .1M sodium propanoate: 4.89

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Most popular questions from this chapter

You have \(75.0 \mathrm{~mL}\) of \(0.10 M\) HA. After adding \(30.0 \mathrm{~mL}\) of \(0.10 M\) \(\mathrm{NaOH}\), the \(\mathrm{pH}\) is \(5.50\). What is the \(K_{\mathrm{u}}\) value of \(\mathrm{HA}\) ?

Amino acids are the building blocks for all proteins in our bodies. A structure for the amino acid alanine is All amino acids have at least two functional groups with acidic or basic properties. In alanine, the carboxylic acid group has \(K_{\mathrm{a}}=4.5 \times 10^{-3}\) and the amino group has \(K_{\mathrm{b}}=7.4 \times 10^{-5} .\) Because of the two groups with acidic or basic properties, three different charged ions of alanine are possible when alanine is dissolved in water. Which of these ions would predominate in a solution with \(\left[\mathrm{H}^{+}\right]=1.0 M ?\) In a solution with \(\left[\mathrm{OH}^{-}\right]=\) \(1.0 M ?\)

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.

Derive an equation analogous to the Henderson-Hasselbalch equation but relating \(\mathrm{pOH}\) and \(\mathrm{p} K_{\mathrm{b}}\) of a buffered solution composed of a weak base and its conjugate acid, such as \(\mathrm{NH}_{3}\) and \(\mathrm{NH}_{4}^{+}\).

Sketch two pH curves, one for the titration of a weak acid with a strong base and one for a strong acid with a strong base. How are they similar? How are they different? Account for the similarities and the differences.
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