Question

Calculate the 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, (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

The pH values for each strong acid solution are: (a) 2.07, (b) 1.38, (c) 1.60, and (d) 0.78.

Step-by-step solution

(a) Calculate H+ concentration for HBr solution

Since the strong acid HBr dissociates fully in water, we have: [H+] = 8.5 × 10⁻³ M

(a) Calculate pH for HBr solution

Now, we can calculate the pH using the formula pH = -log10[H+]: pH = -log10(8.5 × 10⁻³) ≈ 2.07

(b) Calculate H+ concentration for HNO3 solution

First, we need to find the concentration of the HNO3 solution. Using the molar mass of HNO3 (63.01 g/mol), we can calculate the moles of HNO3 present: moles HNO3 = (1.52 g) / (63.01 g/mol) ≈ 0.0241 mol Now, we find the total volume in liters: volume (L) = (575 mL) × (1 L / 1000 mL) = 0.575 L Now, we can find the concentration: [HNO3] = (0.0241 mol) / (0.575 L) ≈ 0.0419 M Since the strong acid HNO3 dissociates fully in water, we have: [H+] = 0.0419 M

(b) Calculate pH for HNO3 solution

Now, we can calculate the pH using the formula pH = -log10[H+]: pH = -log10(0.0419) ≈ 1.38

(c) Calculate H+ concentration for HClO4 solution

First, we need to find the concentration of the HClO4 solution after dilution. The dilution formula is: C₁V₁ = C₂V₂. We have: (0.250 M)(5.00 mL) = (C₂)(50.0 mL) Now, we can solve for C₂: C₂ = (0.250 M)(5.00 mL) / (50.0 mL) = 0.0250 M Since the strong acid HClO4 dissociates fully in water, we have: [H+] = 0.0250 M

(c) Calculate pH for HClO4 solution

Now, we can calculate the pH using the formula pH = -log10[H+]: pH = -log10(0.0250) ≈ 1.60

(d) Calculate H+ concentration for mixed HBr and HCl solution

First, we need to find the moles of H+ contributed by each acid: moles H+ from HBr = (0.100 M)(10.0 mL) = 1.00 mmol moles H+ from HCl = (0.200 M)(20.0 mL) = 4.00 mmol Total moles H+ = 1.00 mmol + 4.00 mmol = 5.00 mmol Now, we find the final volume of the mixed solution (in L): volume (L) = (10.0 mL + 20.0 mL) × (1 L / 1000 mL) = 0.0300 L Now we can find the concentration of H+: [H+] = (5.00 mmol) / (0.0300 L) = 0.1667 M

(d) Calculate pH for mixed HBr and HCl solution

Now, we can calculate the pH using the formula pH = -log10[H+]: pH = -log10(0.1667) ≈ 0.78 In summary, the pH values for each solution are: (a) 2.07, (b) 1.38, (c) 1.60, and (d) 0.78.

Key concepts

Strong Acid Solutions

Strong acid solutions are a fascinating aspect of chemistry! These acids are known for their ability to dissociate completely in water, meaning they break apart into their constituent ions entirely. This complete dissociation is what makes them "strong." When a molecule like HBr or HNO₃ is added to water, it releases all its hydrogen ions (H⁺), resulting in a straightforward calculation of their hydronium ion concentration. With no acid left unionized, you can directly use their initial molarity to determine the concentration of hydrogen ions present in the solution.

Dissociation in Water

Dissociation in water is an essential process in acid-base chemistry. For strong acids, this involves the splitting of the acid molecule into its ions, specifically into positive hydrogen ions (H⁺) and corresponding anions that are part of the acid, like bromide (Br⁻) in the case of HBr. Unlike weak acids, strong acids do not establish an equilibrium but instead dissociate completely as soon as they meet water. This property not only affects their reactivity but makes their pH calculation considerably simpler since their initial concentration can be directly used to calculate the pH.

Hydronium Ion Concentration

The hydronium ion concentration in a solution is pivotal for determining the acidity of the solution. When a strong acid dissociates in water, it releases hydrogen ions which immediately associate with water molecules to form hydronium ions ( H₃O⁺ ). The concentration of hydronium ions is directly equal to the concentration of the dissociating acid in strong acids. For example, if you dissolve 0.0419 M of HNO₃ in water, the concentration of hydronium ions in the solution is also 0.0419 M, simplifying the process of finding the pH because [H⁺] = [H₃O⁺] .

Molarity and Dilution

Understanding molarity and dilution is essential for dealing with acid solutions in chemistry. Molarity, denoted by M, is a measure of concentration, indicating the number of moles of solute per liter of solution. When acids are diluted, their concentration changes according to the formula:$$C_1{V}_1=C_2{V}_2$$ where C_1 and V_1 are the initial concentration and volume, while C_2 and V_2 are the final concentration and volume. This principle helps determine how strong the final solution will be after dilution, allowing for accurate pH calculations post-dilution.

Acid-Base Chemistry

Acid-base chemistry is a broad field studying how acids and bases interact. The strength of an acid like those in the textbook problem lies in its ability to donate hydrogen ions to solutions. Strong acids become crucial as they provide immediate and complete ionization of hydrogen ions, converting them into hydronium. To connect this with pH: any increase in hydronium ions leads to a decrease in the pH value, making solutions more acidic. This direct relationship underpins many reactions and equilibria in chemistry and is crucial for understanding phenomena ranging from biological systems to industrial applications.