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(a) Calculate the pH of a buffer that is \(0.150 \mathrm{M}\) in lactic acid and \(0.120 M\) in sodium lactate. (b) Calculate the pH of a buffer formed by mixing \(75 \mathrm{~mL}\) of \(0.150 \mathrm{M}\) lactic acid with \(25 \mathrm{~mL}\) of \(0.120 \mathrm{M}\) sodium lactate.

Short Answer

Expert verified
The pH of the buffer that is 0.150 M in lactic acid and 0.120 M in sodium lactate is 3.66, and the pH of the buffer formed by mixing 75 mL of 0.150 M lactic acid with 25 mL of 0.120 M sodium lactate is 2.61.

Step by step solution

01

Find pKa for lactic acid

The \(pK_a\) value for lactic acid is 3.86.
02

Step 2(a): Use the Henderson-Hasselbalch equation for 0.150 M lactic acid and 0.120 M sodium lactate

Plug the concentrations into the equation: \(pH = 3.86 + log_{10}\dfrac{0.120}{0.150}\) Now, calculate the pH: \(pH = 3.86 - 0.204\) \(pH = 3.656\)
03

Step 3(a): Round the pH value

Round the pH value to two decimal places to get the pH of the buffer solution: \(pH = 3.66\)
04

Step 2(b): Calculate the new concentrations of lactic acid and sodium lactate by mixing the solutions

Use the given volumes and concentrations to calculate the moles of lactic acid and sodium lactate: Moles of lactic acid = 0.150 M × 0.075 L = 0.01125 mol Moles of sodium lactate = 0.120 M × 0.025 L = 0.00300 mol Now, calculate the new total volume after mixing the solutions: Total volume = 0.075 L + 0.025 L = 0.100 L Determine the new concentrations: [\(HA\)] = (0.01125 mol) / (0.100 L) = 0.1125 M [\(A^-\)] = (0.00300 mol) / (0.100 L) = 0.0300 M
05

Step 3(b): Use the new concentrations in the Henderson-Hasselbalch equation

Plug the new concentrations into the equation: \(pH = 3.86 + log_{10}\dfrac{0.0300}{0.1125}\) Now, calculate the pH: \(pH = 3.86 - 1.25\) \(pH = 2.61\)
06

Step 4(b): Round the pH value

Round the pH value to two decimal places to get the pH of the mixed buffer solution: \(pH = 2.61\) In conclusion, the pH of the buffer that is 0.150 M in lactic acid and 0.120 M in sodium lactate is 3.66, and the pH of the buffer formed by mixing 75 mL of 0.150 M lactic acid with 25 mL of 0.120 M sodium lactate is 2.61.

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Key Concepts

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

Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is a simple yet powerful tool used in chemistry to calculate the pH of a buffer solution. It relates the pH to the pKa (acid dissociation constant) and the concentrations of the acid
  • The equation is: \[pH = pK_a + \log_{10}\left(\frac{[A^-]}{[HA]}\right)\]
  • In this equation:
    • \(pK_a\) is the negative log of the acid dissociation constant, a measure of acid strength
    • \([A^-]\) represents the concentration of the conjugate base
    • \([HA]\) is the concentration of the acid
By using this equation, students are able to calculate the pH of buffer solutions accurately. The beauty of the Henderson-Hasselbalch equation is in its simplicity.
It ties together important concepts such as acid strength and concentration ratios in a way that is easy to grasp.
pH calculation
Calculating the pH of a buffer involves a few straightforward steps once you understand the ingredients of the equation. Let’s walk through:
  • First, identify the acid and base in your buffer system. In this case, lactic acid is the acid and sodium lactate is the conjugate base.
  • Find the pKa value for the acid. For lactic acid, pKa is 3.86.
  • Plug the concentrations of your acid and base into the Henderson-Hasselbalch equation.
    • For instance, if lactic acid concentration is 0.150 M and sodium lactate is 0.120 M, the setup is: \[pH = 3.86 + \log_{10}\left(\frac{0.120}{0.150}\right)\]
    • Calculate the logarithmic term to find the pH.
    • Finally, round off the results to get the desired number of decimal places.
This method allows you to understand the balance between the acid and its conjugate base in a solution, providing insight into how buffering capacities work.
lactic acid buffer
A lactic acid buffer is a type of chemical buffer solution that uses lactic acid and its conjugate base, sodium lactate, to maintain a stable pH level. Buffers are important in many chemical and biological processes because they help resist drastic changes in pH.
Here are some features of the lactic acid buffer:
  • **Components:**
    • Lactic acid (\([HA]\)): An organic acid, which donates \(H^+\) ions in solution.
    • Sodium lactate (\([A^-]\)): It accepts \(H^+\) ions, balancing the \(pH\) by forming lactic acid in return.
  • **Buffer Action:** When a small amount of acid or base is added to the buffer solution, the reaction between \([HA]\) and \([A^-]\) keeps the pH relatively stable.
  • **Applications:** Such a buffer can be utilized in food preservation, sports drinks, and certain pharmaceutical products to ensure they remain effective over a range of conditions.
Using lactic acid buffers helps in maintaining a controlled environment, especially in biological systems where precise pH control is crucial.

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Most popular questions from this chapter

Lead(II) carbonate, \(\mathrm{PbCO}_{3}\), is one of the components of the passivating layer that forms inside lead pipes. (a) If the \(K_{s p}\) for \(\mathrm{PbCO}_{3}\) is \(7.4 \times 10^{-14}\) what is the molarity of \(\mathrm{Pb}^{2+}\) in a saturated solution of lead(II) carbonate? (b) What is the concentration in ppb of \(\mathrm{Pb}^{2+}\) ions in a saturated solution? (c) Will the solubility of \(\mathrm{PbCO}_{3}\) increase or decrease as the \(\mathrm{pH}\) is lowered? (d) The EPA threshold for acceptable levels of lead ions in water is 15 ppb. Does a saturated solution of lead(II) carbonate produce a solution that exceeds the EPA limit?

For each statement, indicate whether it is true or false. (a) The solubility of a slightly soluble salt can be expressed in units of moles per liter. (b) The solubility product of a slightly soluble salt is simply the square of the solubility. (c) The solubility of a slightly soluble salt is independent of the presence of a common ion. (d) The solubility product of a slightly soluble salt is independent of the presence of a common ion.

(a) True or false: "solubility" and "solubility-product constant" are the same number for a given compound. (b) Write the expression for the solubility- product constant for each of the following ionic compounds: \(\mathrm{MnCO}_{3}, \mathrm{Hg}(\mathrm{OH})_{2},\) and \(\mathrm{Cu}_{3}\left(\mathrm{PO}_{4}\right)_{2}\).

(a) The molar solubility of \(\mathrm{PbBr}_{2}\) at \(25^{\circ} \mathrm{C}\) is \(1.0 \times 10^{-2} \mathrm{~mol} / \mathrm{L} .\) Calculate \(K_{s p} .(\mathbf{b})\) If \(0.0490 \mathrm{~g}\) of \(\mathrm{AgIO}_{3}\) dis- solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate \(K_{s p}\) value from Appendix D, calculate the pH of a saturated solution of \(\mathrm{Ca}(\mathrm{OH})_{2}\).

The solubility of two slightly soluble salts of \(\mathrm{M}^{2+}, \mathrm{MA}\) and \(\mathrm{MZ}_{2}\), is the same, \(4 \times 10^{-4} \mathrm{~mol} / \mathrm{L} .(\mathbf{a})\) Which has the larger numerical value for the solubility product constant? (b) In a saturated solution of each salt in water, which has the higher concentration of \(\mathrm{M}^{2+} ?(\mathbf{c})\) If you added an equal volume of a solution saturated in MA to one saturated in \(\mathrm{MZ}_{2}\), what would be the equilibrium concentration of the cation, \(\mathrm{M}^{2+}\) ?

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