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Chapter 14: Problem 64

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 L. Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\) and the pH for this solution. [Hint: Concentrated \(\mathrm{HCl}\) is 38\(\% \mathrm{HCl}\) (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

Expert verified
The concentration of H鈦� ions in the solution is 0.936 M, the concentration of OH鈦� ions is 1.07 脳 10鈦宦光伒 M, and the pH of the solution is approximately 0.03.

Step by step solution

01

Determine the moles of HCl and HNO鈧�

First, we need to find the moles of HCl and HNO鈧� in the solution. We are given the mass percent and densities of both concentrated acids. We will use these information to find the moles of each acid. For HCl: - Mass percent = 38% - Density = 1.19 g/mL - Volume = 50.0 mL mass of HCl = mass percent 脳 density 脳 volume mass of HCl = 0.38 脳 1.19 g/mL 脳 50.0 mL = 22.61 g Now, let's calculate the moles of HCl using its molar mass (36.46 g/mol): moles of HCl = mass of HCl / molar mass of HCl = 22.61 g / 36.46 g/mol = 0.620 mol For HNO鈧�: - Mass percent = 70% - Density = 1.42 g/mL - Volume = 20.0 mL mass of HNO鈧� = mass percent 脳 density 脳 volume mass of HNO鈧� = 0.70 脳 1.42 g/mL 脳 20.0 mL = 19.88 g Now, calculate the moles of HNO鈧� using its molar mass (63.01 g/mol): moles of HNO鈧� = mass of HNO鈧� / molar mass of HNO鈧� = 19.88 g / 63.01 g/mol = 0.316 mol
02

Calculate the total volume of the solution

With the amounts of HCl and HNO鈧� and the water given (300 mL), we also need to add more water until the final volume of the solution is 1.00 L. Total volume of the solution = 1.00 L
03

Calculate the concentration of H鈦� ions

Both HCl and HNO鈧� are strong acids, meaning they completely dissociate in the aqueous solution. The H鈦� ions come from both HCl and HNO鈧�. We add the moles of H鈦� ions from each acid and divide by the total volume (in liters) to find the concentration. Total moles of H鈦� ions = moles of HCl + moles of HNO鈧� = 0.620 mol + 0.316 mol = 0.936 mol Concentration of H鈦� ions [H鈦篯 = total moles of H鈦� ions / total volume of the solution = 0.936 mol / 1.00 L = 0.936 M
04

Find the concentration of OH鈦� ions

We can find the concentration of OH鈦� ions using the ion product of water (Kw): Kw = [H鈦篯[OH鈦籡 = 1.00 脳 10鈦宦光伌 (at 25掳C) We have the [H鈦篯 as calculated in step 3, so we can find the [OH鈦籡: [OH鈦籡 = Kw / [H鈦篯 = (1.00 脳 10鈦宦光伌) / 0.936 M = 1.07 脳 10鈦宦光伒 M
05

Calculate the pH of the solution

Now that we have the concentration of H鈦� ions, we can calculate the pH of the solution: pH = -log[H鈦篯 = -log(0.936) 鈮� 0.03 So, the concentration of H鈦� ions in the solution is 0.936 M, the concentration of OH鈦� ions is 1.07 脳 10鈦宦光伒 M, and the pH of the solution is approximately 0.03.

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Key Concepts

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

Solution Concentration
Solution concentration is a way of expressing how much solute is present in a given volume of solution. In this context, the solutes are hydrochloric acid (HCl) and nitric acid (HNO鈧�).
The concentration can be expressed in different ways, such as molarity, which is often used in chemistry. Molarity (M) is defined as the number of moles of solute per liter of solution.
In the original exercise, we prepared a solution by mixing concentrated acids with water to get a 1-liter solution. The moles of each acid were calculated first, and then their contributions were combined to determine the total concentration of H鈦� ions.
  • Remember that even though the solution includes different acids, the concentration refers to how much of a solute is present per liter in total, not each acid separately.
  • Achieving the desired concentration often involves diluting a more concentrated solution to reach a specific volume with added solvent, such as water.
This understanding of concentration helps predict how the solution will react in chemical processes.
pH Calculation
pH is a measure of the acidity or basicity of a solution. It is calculated using the concentration of hydrogen ions \([H^+]\) in the solution. A solution with a high concentration of \([H^+]\) will have a low pH value, indicating it is acidic.
The formula to find pH is simple: \( ext{pH} = -\log([H^+])\). Using this formula requires you to input the hydrogen ion concentration obtained from the step where you add up the moles from each acid.
  • A low pH value signifies a strong acidic solution, indicating that there are many \([H^+]\) ions present. In our example, a pH of approximately 0.03 indicates a very strong acid.
  • pH is a logarithmic scale, meaning each whole number change on the scale represents a tenfold change in \([H^+]\) concentration.
Thus, the concept of pH is crucial for understanding the behavior of acids and bases in a solution.
Strong Acids
Strong acids, like hydrochloric acid (HCl) and nitric acid (HNO鈧�), are completely dissociated in water. This means that when these acids dissolve in water, they release \([H^+]\) ions entirely, leaving no trace of the original acid molecule in the solution.

Because they dissociate completely, strong acids are particularly effective at lowering the pH of a solution.
  • The original exercise highlights the complete dissociation feature by combining two strong acids to determine the total number of \([H^+]\) ions in the solution.
  • This complete dissociation is why strong acids are treated with caution in lab settings as they significantly alter the solution's properties.
In practical terms, knowing that an acid is strong helps predict its behavior in reactions and its potential to affect the pH of solutions.
Molar Concentration
Molar concentration, or molarity, is a central concept in chemistry, defining the concentration in terms of moles of a solute per liter of solution.
This is particularly useful for solutions that involve reactive components where the number of particles (moles) is more relevant than their weight (mass).

In our exercise, molarity is used to calculate the concentration of both hydrochloric acid and nitric acid once added to the final solution volume. We calculated molarity by taking the total moles of solute and dividing by the solution's volume (in liters).
  • The calculation for molarity is straightforward: \([M] = \frac{\text{moles of solute}}{\text{volume of solution in liters}}\).
  • This measures how much of a substance is present in a specific volume of liquid, providing a basis for understanding solution chemistry.
Understanding molarity makes calculating and working with solutions in labs and chemical equations more manageable.

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Most popular questions from this chapter

For the reaction of hydrazine \(\left(\mathrm{N}_{2} \mathrm{H}_{4}\right)\) in water, $$ \mathrm{H}_{2} \mathrm{NNH}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{NNH}_{3}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ \(K_{\mathrm{b}}\) is \(3.0 \times 10^{-6} .\) Calculate the concentrations of all species and the pH of a \(2.0-M\) solution of hydrazine in water.

A \(1.0 \times 10^{-2}-M\) solution of cyanic acid (HOCN) is 17\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) for cyanic acid.

Will 0.10 M solutions of the following salts be acidic, basic, or neutral? See Appendix 5 for \(K_{\mathrm{a}}\) values. a. ammonium bicarbonate b. sodium dihydrogen phosphate c. sodium hydrogen phosphate d. ammonium dihydrogen phosphate e. ammonium formate

Isocyanic acid \((\mathrm{HNCO})\) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation $$ 2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) $$ Upon isolating pure HNCO \((l),\) an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a 100 -mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) assuming all of the HNCO produced is dissolved in solution? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)

A solution is made by adding 50.0 \(\mathrm{mL}\) of 0.200\(M\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) to 50.0 \(\mathrm{mL}\) of \(1.00 \times 10^{-3} \mathrm{M} \mathrm{HCl}\) a. Calculate the pH of the solution. b. Calculate the acetate ion concentration.
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