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Chapter 14: Problem 55

The pH of a sample of gastric juice in a person's stomach is 2.1. Calculate the pOH, [H'l, and [OH- ] for this sample. Is gastric juice acidic or basic?

Short Answer

Expert verified
The pOH of the gastric juice is 11.9, the [H+] concentration is approximately \(7.94 \times 10^{-3}\) M, and the [OH-] concentration is approximately \(1.26 \times 10^{-12}\) M. Since [H+] > [OH-], gastric juice is acidic.

Step by step solution

01

Calculate pOH

We are given the pH value, which is 2.1. To find the pOH value, we use the relationship between pH and pOH: pH + pOH = 14 Now, we can substitute the given pH value and solve for pOH: 2.1 + pOH = 14 pOH = 14 - 2.1 pOH = 11.9
02

Calculate [H'] concentration

Now, we'll find the [H+] concentration using the given pH value and the formula for pH: pH = -log[H+] We can rearrange the formula to find [H+]: [H+] = 10^(-pH) Substitute the given pH value and solve for [H+]: [H+] = 10^(-2.1) [H+] 鈮� 7.94 脳 10^(-3) M
03

Calculate [OH-] concentration

Next, we will find the [OH-] concentration using the calculated pOH value and the formula for pOH: pOH = -log[OH-] We can rearrange the formula to find [OH-]: [OH-] = 10^(-pOH) Substitute the calculated pOH value and solve for [OH-]: [OH-] = 10^(-11.9) [OH-] 鈮� 1.26 脳 10^(-12) M
04

Determine if gastric juice is acidic or basic

Finally, we'll determine whether the gastric juice is acidic or basic by comparing the concentrations of [H+] and [OH-]. Remember that: - If [H+] > [OH-], then the solution is acidic. - If [H+] < [OH-], then the solution is basic. - If [H+] = [OH-], then the solution is neutral. In our case, the [H+] concentration is 7.94 脳 10^(-3) M, and the [OH-] concentration is 1.26 脳 10^(-12) M. Since [H+] > [OH-], gastric juice is acidic.

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Key Concepts

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

pOH calculation
To calculate the pOH of a solution, we first need to understand its connection to pH. The pH and pOH are related through a simple equation:
\[ \text{pH} + \text{pOH} = 14 \]
This equation implies that the sum of pH and pOH in any solution at 25掳C always equals 14.Knowing this, if we have the pH value, we can easily find the pOH by subtracting the pH from 14.For instance, if the pH is 2.1, the calculation goes as follows:
  • pOH = 14 - pH = 14 - 2.1 = 11.9
This indicates the basic nature of a solution, as a lower pOH corresponds to a higher concentration of OH- ions.
acidic vs. basic solutions
Identifying whether a solution is acidic or basic depends on the concentration of hydrogen ions ([H+]) compared to hydroxide ions ([OH-]).
Here are the key points:
  • If [H+] > [OH-], the solution is acidic.
  • If [H+] < [OH-], the solution is basic.
  • If [H+] = [OH-], the solution is neutral.
For the gastric juice example, we computed the [H+] as approximately\( 7.94 \times 10^{-3} \) M and [OH-] as approximately \( 1.26 \times 10^{-12} \) M.Since [H+] is greater than [OH-], the gastric juice is acidic, which aligns with its typical characteristics, as our stomach's environment favors digestion with acidity.
H+ concentration
The concentration of hydrogen ions in a solution, denoted as [H+], is a key determinant of its acidity. The pH value is a logarithmic measure of [H+] concentration and is calculated using:
\[ \text{pH} = -\log[H^+] \]
To find [H+], the formula can be rearranged:
  • [H+] = 10^{-\text{pH}}
For a pH of 2.1:
  • [H+] = 10^{-2.1} \approx 7.94 \times 10^{-3} \text{ M}
This high concentration of hydrogen ions is typical in acidic solutions, like gastric juice,helping with the breakdown of food during digestion.
OH- concentration
The concentration of hydroxide ions, [OH-], is another critical measure in determining a solution's basicity.
Like pH, pOH provides insight into these concentrations and is logarithmically related to [OH-]:
\[ \text{pOH} = -\log[OH^-] \]
We can derive [OH-] using:
  • [OH-] = 10^{-\text{pOH}}
In the gastric juice scenario with a pOH of 11.9:
  • [OH-] = 10^{-11.9} \approx 1.26 \times 10^{-12} \text{ M}
With such a low concentration of OH- ions, along with the higher concentration of H+, it's clear why gastric juice is classified as acidic.

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Most popular questions from this chapter

A typical sample of vinegar has a pH of \(3.0 .\) Assuming that vinegar is only an aqueous solution of acetic acid \(\left(K_{\mathrm{a}}=1.8 \times\right.\) \(10^{-5}\) ), calculate the concentration of acetic acid in vinegar.

Which of the following conditions indicate a basic solution at \(25^{\circ} \mathrm{C} ?\) a. \(\mathrm{pOH}=11.21\) b. \(\mathrm{pH}=9.42\) c. \(\left[\mathrm{OH}^{-}\right]>\left[\mathrm{H}^{+}\right]\) d. \(\left[\mathrm{OH}^{-}\right]>1.0 \times 10^{-7} \mathrm{M}\)

Consider a solution prepared by mixing a weak acid HA, HCl, and NaA. Which of the following statements best describes what happens? a. The \(\mathrm{H}^{+}\) from the HCl reacts completely with the \(\mathrm{A}^{-}\) from the NaA. Then the HA dissociates somewhat. b. The \(\mathrm{H}^{+}\) from the HCl reacts somewhat with the A- from the NaA to make HA, while the HA is dissociating. Eventually you have equal amounts of everything. c. The \(\mathrm{H}^{+}\) from the HCl reacts somewhat with the \(\mathrm{A}^{-}\) from the NaA to make HA while the HA is dissociating. Eventually all the reactions have equal rates. d. The \(\mathrm{H}^{+}\) from the \(\mathrm{HCl}\) reacts completely with the \(\mathrm{A}^{-}\) from the NaA. Then the HA dissociates somewhat until "too much "H' and \(\mathrm{A}^{-}\) are formed, so the \(\mathrm{H}^{+}\) and \(\mathrm{A}^{-}\) react to form HA, and so on. Eventually equilibrium is reached. Justify your choice, and for choices you did not pick, explain what is wrong with them.

An antacid purchased at a local drug store has a pOH of \(2.3 .\) Calculate the \(\mathrm{pH},\left[\mathrm{H}^{+}\right],\) and \(\left[\mathrm{OH}^{-}\right]\) of this solution. Is the antacid acidic or basic?

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between \(\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\) and \(\left[\mathrm{CO}_{3}^{2-}\right] ?\) c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)
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