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Chapter 13: Problem 56

A solution is prepared by adding \(50.0 \mathrm{mL}\) of \(0.050 \mathrm{M}\) HBr to 150.0 mL of 0.10 \(M\) HI. Calculate \(\left[\mathrm{H}^{+}\right]\) and the pH of this solution. HBr and HI are both considered strong acids.

Short Answer

Expert verified
The concentration of hydrogen ions \(\left(\mathrm{[H^{+}}\right]\) in the solution is \(0.0875\, M\), and the pH of the solution is approximately 1.06.

Step by step solution

01

Calculate moles of HBr and HI

First, we need to determine the moles of HBr and HI in each solution using their molarities and the respective volumes of each solution. The formula to calculate moles is: Moles = Molarity × Volume For HBr: Moles of HBr = \(0.050\, M \times 50.0\, mL = 2.50 \times 10^{-3}\, moles\). For HI: Moles of HI = \(0.10\, M \times 150.0\, mL = 15.0 \times 10^{-3}\, moles\).
02

Calculate the total moles of H⁺ ions

Since both HBr and HI are strong acids and dissociate completely in water, the moles of HBr and HI will contribute to the number of moles of H⁺ ions in the solution. Therefore, the total moles of H⁺ ions are: Moles of H⁺ = Moles of HBr + Moles of HI Moles of H⁺ = \(2.50 \times 10^{-3}\, moles + 15.0 \times 10^{-3}\, moles = 17.5 \times 10^{-3}\, moles\).
03

Calculate the final concentration of H⁺ ions

Next, we need to calculate the resulting concentration of H⁺ ions in the mixed solution. To do this, we'll divide the total moles of H⁺ ions by the total volume of the solution. The total volume of the solution is the sum of the volumes of the individual solutions: Total volume = \(50.0\, mL + 150.0\, mL = 200\, mL\, or \, 0.200\, L\). Now, we can find the concentration of H⁺ ions: \([\mathrm{H^{+}}] = \frac{\text{Total moles of H⁺ ions}}{\text{Total volume of solution}}\). \[\mathrm{[H^{+}]} = \frac{17.5 \times 10^{-3}\, \text{moles}}{0.200\, L}\]. \[\mathrm{[H^{+}]} = 0.0875\, M\].
04

Calculate the pH of the solution

Finally, we can use the formula for pH to calculate the pH value of the solution: pH = - \(\log_{10}\)(H⁺ concentration) pH = - \(\log_{10}(0.0875\, M)\). pH ≈ 1.06 Thus, the concentration of hydrogen ions \(\left(\mathrm{[H^{+}}\right]\) in the solution is \(0.0875\, M\), and the pH of the solution is approximately 1.06.

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Most popular questions from this chapter

At \(25^{\circ} \mathrm{C},\) a saturated solution of benzoic acid \(\left(K_{a}=6.4 \times\right.\) \(10^{-5}\) ) has a pH of \(2.80 .\) Calculate the water solubility of benzoic acid in moles per liter.

Quinine \(\left(\mathrm{C}_{20} \mathrm{H}_{24} \mathrm{N}_{2} \mathrm{O}_{2}\right)\) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, \(\mathrm{p} K_{\mathrm{b}_{1}}=5.1\) and \(\mathrm{p} K_{\mathrm{b}_{2}}=9.7\left(\mathrm{p} K_{\mathrm{b}}=-\log K_{\mathrm{b}}\right) .\) Only 1 g quinine will dissolve in \(1900.0 \mathrm{mL}\) of solution. Calculate the \(\mathrm{pH}\) of a saturated aqueous solution of quinine. Consider only the reaction \(\mathrm{Q}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{QH}^{+}+\mathrm{OH}^{-}\) described by \(\mathrm{p} K_{\mathrm{b}_{1}},\) where \(\mathrm{Q}=\) quinine.

Calculate the percentage of pyridine \(\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)\) that forms pyridinium ion, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+},\) in a \(0.10-M\) aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)

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