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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+] ≈ 7.94 × 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-] ≈ 1.26 × 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 × 10^(-3) M, and the [OH-] concentration is 1.26 × 10^(-12) M. Since [H+] > [OH-], gastric juice is acidic.

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Most popular questions from this chapter

Acrylic acid \(\left(\mathrm{CH}_{2}=\mathrm{CHCO}_{2} \mathrm{H}\right)\) is a precursor for many important plastics. \(K_{\mathrm{a}}\) for acrylic acid is \(5.6 \times 10^{-5} .\) a. Calculate the pH of a \(0.10-M\) solution of acrylic acid. b. Calculate the percent dissociation of a \(0.10-M\) solution of acrylic acid. c. calculate the pH of a \(0.050-M\) solution of sodium acrylate \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Calculate the pH of a \(0.10-M \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is $1.0 \times 10^{-5} .$

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 \(\mathrm{NaHCO}_{3} ?\)

A \(1.0 \times 10^{-2}-M\) solution of cyanic acid (HOCN) is 17\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) for cyanic acid.

Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\mathrm{a}}\) value for hypochlorous acid is \(3.5 \times 10^{-8},\) which is the stronger base, \(\mathrm{OCl}^{-}\) or $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} ?$
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