Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

(a) Calculate the pH of a buffer that is \(0.12 \mathrm{M}\) in lactic acid and \(0.11 M\) in sodium lactate. (b) Calculate the pH of a buffer formed by mixing \(85 \mathrm{~mL}\) of \(0.13 \mathrm{M}\) lactic acid with \(95 \mathrm{~mL}\) of \(0.15 \mathrm{M}\) sodium lactate.

Short Answer

Expert verified
The pH of the first buffer is calculated using the Henderson-Hasselbalch equation to be \(pH = pK_a + \log{\frac{0.11}{0.12}}\), where \(pK_a = -\log_{10}(1.4 \times 10^{-4})\). For the second buffer, the concentrations of lactic acid and sodium lactate after mixing are \(\mathrm{[HA]} = \frac{0.01105}{0.18}\) and \(\mathrm{[A^-]} = \frac{0.01425}{0.18}\), respectively. The pH of this buffer is calculated using the same equation: \(pH = pK_a + \log{\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}}\), where the concentrations are the calculated values.

Step by step solution

01

Finding the pKa of lactic acid

Lactic acid has a chemical formula: \(\mathrm{CH_3CH(OH)COOH}\). The dissociation of lactic acid can be presented by the following equation: \(\mathrm{CH_3CH(OH)COOH} \rightleftharpoons \mathrm{H^+ + CH_3CH(OH)COO^-} \) The acid dissociation constant (K_a) for lactic acid is approximately \(1.4 \times 10^{-4}\). We will use this value to find the pKa of lactic acid: \(pK_a = -\log_{10}{K_a} = -\log_{10}(1.4 \times 10^{-4})\) We can plug this value into a calculator to find out the pKa.
02

Calculate the pH of the first buffer

Now that we have the pKa of lactic acid, we can use the Henderson-Hasselbalch equation to find the pH of the first buffer: \(pH = pK_a + \log{\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}}\) The concentration of lactic acid (HA) is 0.12 M, and the concentration of sodium lactate (A-) is 0.11 M. Plugging these values and the pKa of lactic acid into the equation, we find the pH.
03

Calculate the pH of the second buffer

For the second part of the problem, we need to find the pH of the buffer system formed by mixing volumes of lactic acid and sodium lactate. We will use the given volumes and concentrations to find the overall concentration of lactic acid and sodium lactate in the final buffer and then apply the Henderson-Hasselbalch equation. Amount of lactic acid (in moles) = volume of lactic acid \(\times\) concentration of lactic acid = \((85 \times 10^{-3})L \times 0.13 M = 0.01105\) moles. Amount of sodium lactate (in moles) = volume of sodium lactate \(\times\) concentration of sodium lactate = \((95\times 10^{-3})L \times 0.15 M = 0.01425\) moles. Total volume of the buffer = volume of lactic acid + volume of sodium lactate = \((85 + 95) mL = 180 mL = 0.18 L\) Now we can find the final concentrations in the mixed buffer: Concentration of lactic acid, \(\mathrm{[HA]} = \frac{0.01105 \, \text{moles}}{0.18L}\) Concentration of sodium lactate, \(\mathrm{[A^-]} = \frac{0.01425\, \text{moles}}{0.18L}\) We will use these concentrations to calculate the pH using the Henderson-Hasselbalch equation: \(pH = pK_a + \log{\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}}\) Plug in the pKa of lactic acid and the calculated concentrations for lactic acid, and sodium lactate into the equation to find the pH.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation plays a critical role in buffer chemistry, as it relates the pH of a solution to the pKa and the concentrations of the acid and its conjugate base. In mathematical terms, the equation is expressed as follows:\[pH = pK_a + \log{\frac{[A^-]}{[HA]}} \]In this equation:
  • \([HA]\) represents the concentration of the acid (in this case, lactic acid).
  • \([A^-]\) represents the concentration of the conjugate base (sodium lactate).
Using this equation, you can calculate the pH of a buffer system if you know the pKa of the acid and the concentrations of the acid and base in the solution. It shows that the pH is closest to the pKa value when the concentrations of the acid and its base are similar. This balance makes buffers effective at resisting changes in pH when acids or bases are added.
pH calculation
Calculating pH using the Henderson-Hasselbalch equation involves several steps. First, you need the pKa value of the acid.
For lactic acid, this comes from its dissociation constant, which is a measure of the acid's strength. Knowing the pKa is crucial, as it is directly used in the calculation.
Once you have the pKa, you calculate log ratio of the concentrations of the conjugate base to the acid. Evaluate as follows:\[pH = pK_a + \log{\frac{[A^-]}{[HA]}} \]Ensure you substitute the correct values into the equation:
  • Use the accurate concentration values for the calculations. For example, if the solution contains 0.12 M lactic acid and 0.11 M sodium lactate, replace \([HA]\) with 0.12 and \([A^-]\) with 0.11.
  • Carry out the calculation to receive the resulting pH.
Understanding this process is essential, especially when working with buffer problems in chemistry, because it allows for the maintenance of pH in a narrow range, crucial in many biological and chemical processes.
lactic acid dissociation
Lactic acid is a common organic acid with the formula \(\mathrm{CH_3CH(OH)COOH}\). When dissolved in water, it can lose a proton to form its conjugate base, lactate, according to the equation:\[\mathrm{CH_3CH(OH)COOH} \rightleftharpoons \mathrm{H^+} + \mathrm{CH_3CH(OH)COO^-}\]This process is an example of acid dissociation, where the acid releases a proton (\(\mathrm{H^+}\)) into the solution, increasing the hydrogen ion concentration.
The extent of dissociation of an acid is quantified by its dissociation constant (\(K_a\)), which for lactic acid is about \(1.4 \times 10^{-4}\). This value indicates that lactic acid is a weak acid since it does not completely dissociate in water.
To find the pKa, which is often more useful in various calculations, take the negative logarithm of \(K_a\):\[pK_a = -\log_{10}(K_a)\]This results in a pKa value around 3.85, representing a crucial part of determining the pH of buffer solutions containing lactic acid.
acid-base chemistry
Acid-base chemistry is fundamental for understanding buffers, which help maintain pH stability in solutions. The basic concept involves acids (proton donors) and bases (proton acceptors).
When you combine these in a buffer, you form a solution that resists pH changes when small amounts of acids or bases are added. This is due to the equilibrium formed between the acid and its conjugate base.

For example, in the buffering system involving lactic acid and sodium lactate:
  • Lactic acid donates protons to form lactate.
  • The sodium lactate acts as a base, accepting protons.
This dynamic creates a buffer solution, which is very useful in conditions where maintaining a constant pH is essential, such as in physiological processes or certain industrial applications. Understanding how the equilibrium shifts in response to added acids or bases is key to mastering acid-base chemistry and buffer solution calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The solubility product for \(\mathrm{Zn}(\mathrm{OH})_{2}\) is \(3.0 \times 10^{-16} .\) The formation constant for the hydroxo complex, \(\mathrm{Zn}(\mathrm{OH})_{4}^{2-},\) is \(4.6 \times 10^{17}\). What concentration of \(\mathrm{OH}^{-}\) is required to dissolve 0.015 mol of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in a liter of solution?

A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}\) and \(0.010 \mathrm{M}\) in \(\mathrm{Sr}^{2+}\). (a) What con- centration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4}: K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}:\) \(\left.K_{s p}=3.2 \times 10^{-7} .\right)\) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}\) when the second cation begins to precipitate?

Predict whether the equivalence point of each of the following titrations is below, above, or at \(\mathrm{pH} 7:\) (a) formic acid titrated with \(\mathrm{NaOH},\) (b) calcium hydroxide titrated with perchloric acid, (c) pyridine titrated with nitric acid.

How many milliliters of \(0.0850 \mathrm{M} \mathrm{NaOH}\) are required to titrate each of the following solutions to the equivalence point: (a) \(40.0 \mathrm{~mL}\) of \(0.0900 \mathrm{M} \mathrm{HNO}_{3}\), (b) \(35.0 \mathrm{~mL}\) of \(0.0850 \mathrm{M}\) \(\mathrm{CH}_{3} \mathrm{COOH},\) (c) \(50.0 \mathrm{~mL}\) of a solution that contains \(1.85 \mathrm{~g}\) of HCl per liter?

(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubility-product constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: AgI, \(\mathrm{SrSO}_{4}, \mathrm{Fe}(\mathrm{OH})_{2},\) and \(\mathrm{Hg}_{2} \mathrm{Br}_{2}\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free