Question

A solution is prepared by adding \(50.0 \mathrm{~mL}\) concentrated hydrochloric acid and \(20.0 \mathrm{~mL}\) concentrated nitric acid to \(300 \mathrm{~mL}\) water. More water is added until the final volume is \(1.00 \mathrm{~L}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) for this solution. [Hint: Concentrated HCl is \(38 \%\) HCl (by mass) and has a density of \(1.19\) \(\mathrm{g} / \mathrm{mL} ;\) concentrated \(\mathrm{HNO}_{3}\) is \(70 . \% \mathrm{HNO}_{3}\) (by mass) and has a density of \(1.42 \mathrm{~g} / \mathrm{mL} .\) ]

Short answer

In the given solution, the concentration of H+ ions is \([H^+] = 0.9356 \mathrm{M}\), the concentration of OH- ions is \([OH^-] = 1.0686 \times 10^{-14} \mathrm{M}\), and the pH of the solution is \(0.0292\).

Step-by-step solution

Find the moles of HCl and HNO3 in the solution

To do this, we will use the given concentration (percentage by mass) and the density of concentrated HCl and HNO3. Moles of HCl: First, let's find the mass of HCl present in 50mL concentrated HCl: Mass of Solution = Volume x Density = 50 mL x 1.19 g/mL = 59.5 g Mass of HCl = (Percentage by mass) x (Mass of solution)/100 = (38%) x (59.5 g) / 100 = 22.61 g Now, moles of HCl = Mass of HCl/Molar_mass Molar_mass of HCl = 36.461 g/mol Moles of HCl = 22.61 g / 36.461 g/mol = 0.6199 mol HCl Moles of HNO3: Mass of Nitric acid Solution = Volume x Density = 20 mL x 1.42 g/mL = 28.4 g Mass of HNO3 = (Percentage by mass) x (Mass of solution)/100 = (70%) x (28.4 g) / 100 = 19.88 g Now, moles of HNO3 = Mass of HNO3/Molar_mass Molar_mass of HNO3 = 63.01 g/mol Moles of HNO3 = 19.88 g / 63.01 g/mol = 0.3157 mol HNO3

Find the concentration of H+ ions in the solution

Now, let's find the total moles of H+ ions: Total moles of H+ ions = Moles of HCl + Moles of HNO3 = 0.6199 mol + 0.3157 mol = 0.9356 mol The final volume of the solution is given as 1 L. So, the concentration of H+ ions is: [H+] = Total moles of H+ ions/Volume = 0.9356 mol / 1 L = 0.9356 M

Calculate pH

Now that we have the concentration of H+ ions, we can calculate the pH of the solution as follows: pH = -log10([H+]) = -log10(0.9356) = 0.0292

Calculate the concentration of OH- ions

We can use the formula for the ion product of water: Kw = [H+]*[OH-] = 1.0 x 10^(-14) Now, we can find [OH-] using the [H+] we calculated in step 2: [OH-] = Kw/[H+] = (1.0 x 10^(-14)) / (0.9356 M) = 1.0686 x 10^(-14) M The final values are as follows: [H+] = 0.9356 M [OH-] = 1.0686 x 10^(-14) M pH = 0.0292

Key concepts

Concentration of Hydrogen Ions

Understanding the concentration of hydrogen ions, often represented as \[H^+\], is crucial in determining the acidic properties of a solution. These ions are protons that have been released by acids when they dissolve in water. In the case of the mixture of hydrochloric acid (HCl) and nitric acid (HNO₃), both contribute to the \[H^+\] concentration in the solution.

When we calculate the concentration of these ions, we are effectively determining the strength of the acid solution. The 'concentration' refers to the amount of substance per unit volume, and this is typically expressed in molarity (M), which is moles per liter. In the given problem, the total volume of the solution is 1.00 L, and it is crucial to consider the volume of the solvent added along with the acids' contributions, to get the accurate molarity of the hydrogen ions. This will be the sum of moles of hydrogen ions from both HCl and HNO₃ divided by the total volume of the solution.

pH Calculation

The pH is a logarithmic scale used to measure the acidity or alkalinity of a solution. It is calculated as the negative logarithm (base 10) of the hydrogen ion concentration. Calculating the pH from the molarity of hydrogen ions helps us understand how acidic or basic a solution is without directly handling potentially hazardous chemicals.

The formula used for calculating the pH from the hydrogen ion concentration is given by \[ pH = -\log_{10}\left([H^+]\right) \.\] The process involves taking the negative logarithm of the \[H^+\] molarity. As the concentration of \[H^+\] increases, the pH value decreases, indicating a more acidic solution. This exponential relationship means that each whole number pH change represents a tenfold change in hydrogen ion concentration, making pH a very sensitive scale.

Stoichiometry in Solutions

Stoichiometry in solutions involves calculations based on the relationships between the amounts of reactants and products in a chemical reaction. In the context of acid and base reactions, stoichiometry helps us determine the number of moles of hydrogen or hydroxide ions produced. For instance, HCl dissociates completely in water, releasing one mole of \[H^+\] ions for every mole of HCl. In contrast, some acids or bases may release more or fewer ions depending on their chemical formula.

In our exercise, the correct use of stoichiometry is essential for finding the exact moles of each acid based on their mass-percent concentration and density. After calculating the moles of the acids, we then infer the amount of \[H^+\] ions produced. As both HCl and HNO₃ release one \[H^+\] ion per molecule, the stoichiometry is straightforward: the moles of acid directly translate to moles of \[H^+\] ions.

Molarity of Acids

The molarity of an acid solution, often abbreviated as 'M,' is a measure of the concentration of acid in a solution, specifically referring to the number of moles of acid per liter of solution. It is one of the most fundamental concepts in chemistry for quantifying the strength of an acid. Molarity is an invaluable tool for scientists because it allows them to predict how an acid will react with other substances and the outcomes of chemical reactions.

For our exercise, to find the initial molarity of the acids, we need the density of the concentrated acids along with the mass percentage of the acid in the solution. Using these, we can calculate the number of moles of acid present in the given volumes and thus their molarity. When diluting these acids to make a 1.00 L solution, it's important to consider the dilution factor in calculating the new molarity of the acids, which is a key component in the pH calculation for the diluted solution.