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Chapter 16: Problem 118

The volume of an adult's stomach ranges from about 50 \(\mathrm{mL}\) when empty to \(1 \mathrm{~L}\) when full. If the stomach volume is \(400 \mathrm{~mL}\) and its contents have a pH of 2 , how many moles of \(\mathrm{H}^{+}\) does the stomach contain? Assuming that all the \(\mathrm{H}^{+}\) comes from \(\mathrm{HCl}\), how many grams of sodium hydrogen carbonate will totally neutralize the stomach acid?

Short Answer

Expert verified
The concentration of H+ ions is given by \( [H^+] = 10^{-2} \mathrm{M} \). Then, the number of moles of H+ ions in the 400 mL solution is \( n_{H^{+}} = 10^{-2} \mathrm{M} \times 0.4 \mathrm{L} = 0.004 \, \mathrm{moles} \). Since the reaction between H+ ions and sodium hydrogen carbonate (NaHCO3) occurs in a 1:1 molar ratio, \( n_{NaHCO3} = n_{H^{+}} = 0.004 \, \mathrm{moles} \). Finally, the mass of sodium hydrogen carbonate needed for neutralization is \( m_{NaHCO3} = n_{NaHCO3} \times M_{NaHCO3} = 0.004\, \mathrm{moles} \times 84\,\mathrm{g/mol} = 0.336\, \mathrm{g} \). Therefore, 0.336 grams of sodium hydrogen carbonate are needed to completely neutralize the stomach acid.

Step by step solution

01

Find the concentration of H+ ions

Using the pH formula, we have: pH = -log(\( [H^+] \)), where \( [H^+] \) represents the concentration of H+ ions. Given the pH value as 2, we can find the concentration of H+ ions as follows: \[2 = -\log{[H^{+}]}\]
02

Calculate the number of moles of H+ ions in the 400mL solution

Now, we have to find the number of moles of H+ in the 400mL stomach solution. First, we need to find the concentration of H+ ions in the solution: \[ [H^+] = 10^{-2} \mathrm{M} \] Next, we can calculate the number of moles of H+ ions (n) using the formula: n = \( [H^+] \) × Volume where n represents the number of moles, and the volume of the solution should be in liters. So, the number of moles of H+ ions is: \[ n = 10^{-2} \mathrm{M} \times 0.4 \mathrm{L} \]
03

Determine the moles of sodium hydrogen carbonate needed for neutralization

Sodium hydrogen carbonate (NaHCO3) reacts with HCl in a 1:1 molar ratio. Therefore, the number of moles of NaHCO3 needed to neutralize the stomach acid is equal to the number of moles of H+ ions calculated in Step 2: \[ n_{NaHCO3} = n_{H^{+}} \]
04

Calculate the mass of sodium hydrogen carbonate needed for neutralization

Now, to find the mass of sodium hydrogen carbonate (NaHCO3), we need to multiply the number of moles of NaHCO3 by its molar mass. The molar mass of NaHCO3 is approximately 84 g/mol. \[ m_{NaHCO3} = n_{NaHCO3} \times M_{NaHCO3} \] Finally, substituting the values: \[ m_{NaHCO3} = (10^{-2} \mathrm{M} \times0.4\mathrm{L}) \times 84\; \mathrm{g/mol} \] Now, your task is to plug in the values and do the math in all the steps to find the mass of sodium hydrogen carbonate needed to neutralize the stomach acid.

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Most popular questions from this chapter

(a) Write a chemical equation that illustrates the autoionization of water. (b) Write the expression for the ionproduct constant for water, $K_{w} .(\mathbf{c})$ If a solution is described as basic, which of the following is true: (i) \(\left[\mathrm{H}^{+}\right]>\left[\mathrm{OH}^{-}\right]\), (ii) \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right],\) or (iii) \(\left[\mathrm{H}^{+}\right]<\left[\mathrm{OH}^{-}\right] ?\)

Calculate \(\left[\mathrm{OH}^{-}\right]\) and \(\mathrm{pH}\) for each of the following strong base solutions: (a) \(0.182 \mathrm{M} \mathrm{KOH},\) (b) \(3.165 \mathrm{~g}\) of \(\mathrm{KOH}\) in 500.0 \(\mathrm{mL}\) of solution, \((\mathbf{c}) 10.0 \mathrm{~mL}\) of $0.0105 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_{2}\( diluted to \)500.0 \mathrm{~mL},(\mathbf{d})\( a solution formed by mixing \)20.0 \mathrm{~mL}$ of 0.015 \(M \mathrm{Ba}(\mathrm{OH})_{2}\) with \(40.0 \mathrm{~mL}\) of $8.2 \times 10^{-3} \mathrm{M} \mathrm{NaOH}$.

(a) Give the conjugate base of the following Brønsted Lowry acids: (i) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-},\) (ii) HBr. (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) \(\mathrm{CN}^{-},\) (ii) \(\mathrm{HSO}_{4}^{-}\).

Write the chemical equation and the \(K_{b}\) expression for the reaction of each of the following bases with water: (a) trimethylamine, \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} ;\) (b) sulfite, \(\mathrm{SO}_{3}^{2-}\); (c) cyanide, \(\mathrm{CN}^{-}\).

(a) Give the conjugate base of the following BrønstedLowry acids: (i) \(\mathrm{H}_{2} \mathrm{SO}_{3},\) (ii) \(\mathrm{HSO}_{3}^{-}\) (b) Give the conjugate acid of the following Bronsted-Lowry bases: (i) $\mathrm{CH}_{3} \mathrm{NH}_{2}$, (ii) \(\mathrm{CH}_{3} \mathrm{COO}^{-}\).
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