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Calculate the \(\mathrm{pH}\) of each of the following strong acid solutions: (a) \(8.5 \times 10^{-3} \mathrm{M} \mathrm{HBr}\), (b) \(1.52 \mathrm{~g}\) of \(\mathrm{HNO}_{3}\) in \(575 \mathrm{~mL}\) of solution, \((\mathrm{c}) 5.00 \mathrm{~mL}\) of \(0.250 \mathrm{M} \mathrm{HClO}_{4}\) diluted to \(50.0 \mathrm{~mL}\), (d) a solution formed by mixing \(10.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HBr}\) with \(20.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{HCl}\)

Short Answer

Expert verified
The calculated pH values for the given strong acid solutions are: (a) 2.07 for 8.5x10鈦宦 M HBr, (b) 1.38 for 1.52 g HNO鈧 in 575 mL of solution, (c) 1.60 for 5.00 mL of 0.250 M HClO鈧 diluted to 50.0 mL, and (d) 0.78 for the solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl.

Step by step solution

01

(a) Calculate the hydronium ion concentration for 8.5x10鈦宦 M HBr

Since HBr is a strong acid, it will completely dissociate in water: HBr 鈫 H鈦 + Br鈦. Therefore, the concentration of HBr is equal to the hydronium ion concentration: [H鈧僌鈦篯 = 8.5x10鈦宦 M.
02

(a) Find the pH of the 8.5x10鈦宦 M HBr solution

Now use the pH formula: pH = -log[H鈧僌鈦篯 = -log(8.5x10鈦宦) 鈮 2.07.
03

(b) Calculate the hydronium ion concentration for 1.52 g HNO鈧 in 575 mL of solution

First, find the moles of HNO鈧: moles = mass/molar mass, where the molar mass of HNO鈧 鈮 63 g/mol, so moles = 1.52 g / 63 g/mol 鈮 0.0241 mol. Now, find the molarity: [HNO鈧僝 = moles/volume = 0.0241 mol / 0.575 L 鈮 0.042 M. Since HNO鈧 is a strong acid, [H鈧僌鈦篯 = [HNO鈧僝 = 0.042 M.
04

(b) Find the pH of the 1.52 g HNO鈧 in 575 mL of solution

Use the pH formula: pH = -log[H鈧僌鈦篯 = -log(0.042) 鈮 1.38.
05

(c) Calculate the hydronium ion concentration for 5.00 mL of 0.250 M HClO鈧 diluted to 50.0 mL

Use the dilution formula: M1V1 = M2V2, where M1 = 0.250 M, V1 = 0.005 L, and V2 = 0.050 L. Solve for M2: M2 = M1V1 / V2 = (0.250 M)(0.005 L) / 0.050 L 鈮 0.025 M. Since HClO鈧 is a strong acid, [H鈧僌鈦篯 = [HClO鈧刔 = 0.025 M.
06

(c) Find the pH of the 5.00 mL of 0.250 M HClO鈧 diluted to 50.0 mL

Use the pH formula: pH = -log[H鈧僌鈦篯 = -log(0.025) 鈮 1.60.
07

(d) Calculate the hydronium ion concentration for the solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl

Find the moles of HBr: moles = M x V = 0.100 M x 0.010 L = 0.001 mol. Find the moles of HCl: moles = M x V = 0.200 M x 0.020 L = 0.004 mol. Combine the moles of both acids: total moles = 0.001 mol + 0.004 mol = 0.005 mol. Now, find the total volume of the mixed solution: total volume = 0.010 L + 0.020 L = 0.030 L. Calculate the hydronium ion concentration: [H鈧僌鈦篯 = total moles / total volume = 0.005 mol / 0.030 L 鈮 0.167 M.
08

(d) Find the pH of the solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCl

Use the pH formula: pH = -log[H鈧僌鈦篯 = -log(0.167) 鈮 0.78.

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Key Concepts

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

Strong Acid Dissociation
Understanding how strong acids dissociate in water is crucial for predicting the pH of a solution. Strong acids, such as hydrobromic acid (HBr) and nitric acid (HNO鈧), completely dissociate into their ions in water. This means that every molecule of a strong acid will break apart to release a hydronium ion (H鈧僌鈦) and its conjugate base (e.g., Br鈦 for HBr).

The dissociation equation for a strong acid like HBr would be as follows:
\[ \text{HBr} \rightarrow \text{H}^{+} + \text{Br}^{-} \].

Since the dissociation is complete, the molarity of the acid solution is equal to the concentration of hydronium ions. In exercises dealing with strong acids, you can directly equate the acid's initial molarity to the hydronium ion concentration for pH calculations.
Hydronium Ion Concentration
When strong acids dissolve in water, they release hydronium ions (H鈧僌鈦), which are responsible for the acidic properties of the solution. The concentration of these ions is expressed in moles per liter (M). The pH of an aqueous solution is the negative logarithm (base 10) of the hydronium ion concentration:
\[ \text{pH} = -\log[H_3O^+] \].

To calculate the hydronium ion concentration from a given molarity of a strong acid, we can use the stoichiometry of the acid's dissociation reaction. For example, in the case of HBr with a molarity of \(8.5 \times 10^{-3} \text{M}\), the concentration of H鈧僌鈦 is the same. In dilution problems, the concentration of hydronium ions changes according to the dilution factor and needs to be recalculated before determining the pH.
pH Formula Application
The pH scale is a measure of the acidity or basicity of an aqueous solution. The pH is calculated using the formula:
\[ \text{pH} = -\log[H_3O^+] \].

For strong acids that dissociate completely, the initial molarity of the acid translates directly to the concentration of H鈧僌鈦 ions. Once we determine the hydronium ion concentration, applying the pH formula is straightforward. Logarithms can be challenging, but remember that a higher concentration of hydronium ions results in a lower pH, making the solution more acidic. For instance, a hydronium ion concentration of \(0.042 \text{M}\) for HNO鈧 results in a pH of approximately 1.38, reflecting a highly acidic solution.
Acid Dilution Calculations
Dilution is the process of reducing the concentration of a solute in solution, usually by adding more solvent. The key principle in any dilution is that the amount of solute remains constant before and after the dilution process:
\[ \text{Moles before dilution} = \text{Moles after dilution} \].
This can be translated to molarity and volume using the formula: \[ \text{M}_1 \times \text{V}_1 = \text{M}_2 \times \text{V}_2 \], where \( \text{M}_1 \) and \( \text{V}_1 \) are the molarity and volume before dilution, and \( \text{M}_2 \) and \( \text{V}_2 \) are the molarity and volume after dilution. For example, in the case of HClO鈧 diluted from 5.00 mL to 50.0 mL, the initial concentration is reduced, impacting the calculation of hydronium ion concentration and resulting pH.

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Most popular questions from this chapter

The active ingredient in aspirin is acetylsalicylic acid \(\left(\mathrm{HC}_{9} \mathrm{H}, \mathrm{O}_{4}\right)\), a monoprotic acid with \(K_{a}=3.3 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\). What is the \(\mathrm{pH}\) of a solution obtained by dissolving two extra-strength aspirin tablets, containing \(500 \mathrm{mg}\) of acetylsalicylic acid each, in \(250 \mathrm{~mL}\) of water?

Designate the Br酶nsted-Lowry acid and the BronstedLowry base on the left side of each of the following equations, and also designate the conjugate acid and conjugate base on the right side: (a) \(\mathrm{NH}_{4}{ }^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{NH}_{3}(a q)\) (b) \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N}(a q)+\mathrm{H}_{2} \mathrm{O}(I) \rightleftharpoons\) (c) \(\mathrm{HCHO}_{2}(a q)+\mathrm{PO}_{4}{ }^{3-}(a q) \stackrel{\left(\mathrm{CH}_{3}\right)_{3} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)}{\mathrm{CHO}_{2}^{-}(a q)+\mathrm{HPO}_{4}{ }^{2-}(a q)}\)

Ephedrine, a central nervous system stimulant, is used in nasal sprays as a decongestant. This compound is a weak organic base: \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}(a q)+\mathrm{OH}^{-}(a q)\) A \(0.035 \mathrm{M}\) solution of ephedrine has a pH of \(11.33 .\) (a) What are the equilibrium concentrations of \(\mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ON}, \mathrm{C}_{10} \mathrm{H}_{15} \mathrm{ONH}^{+}\), and \(\mathrm{OH}^{-} ?\) (b) Calculate \(K_{b}\) for ephedrine.

Explain the following observations: (a) \(\mathrm{HNO}_{3}\) is a stronger acid than \(\mathrm{HNO}_{2} ;\) (b) \(\mathrm{H}_{2} \mathrm{~S}\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{O} ;\) (c) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger acid than \(\mathrm{HSO}_{4}^{-} ;\) (d) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger acid than \(\mathrm{H}_{2} \mathrm{SeO}_{4} ;\) (e) \(\mathrm{CCl}_{3} \mathrm{COOH}\) is a stronger acid than \(\mathrm{CH}_{3} \mathrm{COOH}\)

(a) What is a strong base? (b) A solution is labeled \(0.035 \mathrm{M} \mathrm{Sr}(\mathrm{O} \mathrm{H})_{2}\). What is \(\left[\mathrm{OH}^{-}\right]\) for the solution? (c) Is the following statement true or false? Because \(\mathrm{Mg}(\mathrm{OH})_{2}\) is not very soluble, it cannot be a strong base. Explain.

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