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Chapter 14: Problem 58

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

Short Answer

Expert verified
The concentration of H⁺ ions in the mixed solution is 0.0875 M, and the pH of the solution is 1.06.

Step by step solution

01

Identifying the data from the problem

Let's first list down all given information and formulas we are going to use. - Volume of HBr solution: 50.0 mL - Concentration of HBr solution: 0.050 M - Volume of HI solution: 150.0 mL - Concentration of HI solution: 0.10 M - We need to find: [H⁺] and pH Since both HBr and HI are strong acids, they dissociate completely in water: \(HBr \rightarrow H^{+} + Br^{-}\) \(HI \rightarrow H^{+} + I^{-}\)
02

Convert volumes to liters

We need the volumes in liters for our calculations. We can convert the given volumes in milliliters (mL) to liters (L) by the following conversions: 1 L = 1000 mL For 50.0 mL of HBr solution: Volume = 50.0 mL * (1 L / 1000 mL) = 0.050 L For 150.0 mL of HI solution: Volume = 150.0 mL * (1 L / 1000 mL) = 0.150 L
03

Calculate the total moles of H⁺ produced from both acids

Next, we calculate the moles of H⁺ ions produced by both HBr and HI solutions using the given concentrations and volumes using the formula: moles of H⁺ = concentration × volume For HBr: moles of H⁺ from HBr = 0.050 M * 0.050 L = 0.0025 mol For HI: moles of H⁺ from HI = 0.10 M * 0.150 L = 0.015 mol Now, find the total moles of H⁺ produced by adding the moles from both acids: Total moles of H⁺ = moles of H⁺ from HBr + moles of H⁺ from HI = 0.0025 mol + 0.015 mol = 0.0175 mol
04

Calculate the final concentration of H⁺ in the solution

The total volume of the solution is the sum of the volumes of HBr and HI solutions: Total volume = volume of HBr + volume of HI = 0.050 L + 0.150 L = 0.200 L Now, calculate the final concentration of H⁺ ions in the solution using the formula: Concentration of H⁺ = (total moles of H⁺) / (total volume of the solution) [H⁺] = 0.0175 mol / 0.200 L = 0.0875 M
05

Calculate the pH of the solution

Now that we have the concentration of H⁺ ions, we can calculate the pH using the formula: pH = -log[H⁺] pH = -log(0.0875) = 1.06 Thus, the concentration of H⁺ ions in the mixed solution is 0.0875 M and the pH of the solution is 1.06.

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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 \(\left(\mathrm{NaC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

Saccharin, a sugar substitute, has the formula $\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{NSO}_{3}$ and is a weak acid with \(K_{\mathrm{a}}=2.0 \times 10^{-12} .\) If 100.0 \(\mathrm{g}\) of saccharin is dissolved in enough water to make 340 \(\mathrm{mL}\) of solution, calculate the \(\mathrm{pH}\) of the resulting solution.

A \(0.15-M\) solution of a weak acid is 3.0\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) .

For the following, mix equal volumes of one solution from Group I with one solution from Group II to achieve the indicated pH. Calculate the pH of each solution. $$\begin{aligned} \text { Group I: } & 0.20 M \mathrm{NH}_{4} \mathrm{Cl}, 0.20 \mathrm{MCl}, 0.20 M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl} \\ & 0.20 M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{3} \mathrm{NHCl} \end{aligned}$$ $$\begin{aligned} \text { Group II: } 0.20 \quad M\quad \mathrm{KOI}, 0.20 \quad\mathrm{M} \quad\mathrm{NaCN}, 0.20\quad \mathrm{M}\quad \mathrm{KOCl}, 0.20 \\ \mathrm{M}\quad \mathrm{NaNO}_{2} \end{aligned}$$ a. the solution with the lowest pH b. the solution with the highest pH c. the solution with the pH closest to 7.00

Calculate the concentration of an aqueous \(\mathrm{Sr}(\mathrm{OH})_{2}\) that has \(\mathrm{pH}=10.50\) .
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