Question

Calculate the pH of the following solutions: a. 1.2\(M \mathrm{CaBr}_{2}\) b. 0.84\(M \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}\left(K_{\mathrm{b}} \text { for } \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}=3.8 \times 10^{-10}\right)\) c. 0.57\(M \mathrm{KC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}} \text { for } \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}=6.4 \times 10^{-5}\right)\)

Short answer

a. The pH of 1.2 M CaBr2 solution is 7.00. b. The pH of 0.84 M C_6H_5NH_3NO_3 solution is 0.075. c. The pH of 0.57 M KC_7H_5O_2 solution can be calculated using the given Ka value and the formula pH = 14 - pOH.

Step-by-step solution

Identify the nature of the solution

Determine if the solution is acidic, basic, or neutral. This will depend on the nature of the compound present in the solution. a. CaBr2 is a salt derived from a strong base (Ca(OH)2) and a strong acid (HBr), hence the solution will be neutral. b. C6H5NH3NO3 is derived from a weak base (C6H5NH2) and a strong acid (HNO3), hence the solution will be acidic. c. KC7H5O2 is derived from a strong base (KOH) and a weak acid (HC7H5O2), hence the solution will be basic.

Calculate H+ or OH- concentrations

Calculate the molar concentrations of H+ or OH- ions depending on whether the compound is acidic or basic. a. For CaBr2, since it is a neutral salt, there is no need to calculate concentrations of H+ or OH- as [H+] = [OH-] = 10^-7 M. b. For C6H5NH3NO3, the acidic solution will have equal concentration of [H+]: 0.84M c. For KC7H5O2, calculate OH- concentration using Kb Kb = ([OH-][C7H5O-])/[HC7H5O2] 6.4*10^-5 = (x *0.57)/(1 - x) Calculate x

Calculate pH and pOH

Calculate pH or pOH of the solutions based on whether they contain H+ or OH- ions. a. For CaBr2, [H+] = 10^-7 M, pH = -log(10^-7) = 7.00 b. For C6H5NH3NO3, pH = -log(0.84) ≈ 0.075 c. For KC7H5O2, calculate pOH using [OH-], and then calculate pH. pOH = -log(x) pH = 14 - pOH

Report the pH values

Provide the pH values of the given solutions. a. The pH of 1.2 M CaBr2 solution is 7.00. b. The pH of 0.84 M C6H5NH3NO3 solution is 0.075. c. The pH of 0.57 M KC7H5O2 solution can be obtained using the calculations in step 3.

Key concepts

Acid-Base Chemistry

Acid-base chemistry revolves around the concept of acids and bases, which are substances that can donate or accept protons, respectively. Understanding whether a compound is an acid or a base is essential for predicting its behavior in solution.
An acid is a substance that increases the concentration of hydrogen ions ( H^+ ) when dissolved in water. Common examples include hydrochloric acid (HCl) or nitric acid (HNO3). On the other hand, a base increases the concentration of hydroxide ions ( OH^- ) when in solution, such as sodium hydroxide (NaOH).

• Strong acids and bases dissociate completely in water.
• Weak acids and bases only partially dissociate.

Salts can also affect the pH of a solution depending on their origins. For example, a salt formed from a strong acid and a strong base tends to be neutral, like CaBr2.
Meanwhile, a salt from a weak base and strong acid typically results in an acidic solution, like C6H5NH3NO3.

Molarity

Molarity is a measure of the concentration of a solution. It is defined as the number of moles of solute divided by the volume of the solution in liters. It's commonly used in chemistry to express concentrations precisely.
Understanding molarity is crucial for calculating the amount of a substance present in a given volume of solution, which in turn affects reactions and pH calculations. For example, the molarity of C6H5NH3NO3 is given as 0.84 M, meaning there are 0.84 moles of the compound dissolved per liter of solution.

• Formula: Molarity (M) = moles of solute / liters of solution.
• It helps in determining the concentration of H^+ (or OH^- ) ions in solutions which is crucial for pH calculations.

By using molarity, one can accurately compute other important properties like equilibrium concentration and pH of the solution.

Equilibrium Constant

The equilibrium constant ( K ) is a value that expresses the ratio of products to reactants at equilibrium for a reversible reaction. In acid-base chemistry, two specific equilibrium constants are often considered: Ka for acids, and Kb for bases.
These constants provide a way to quantify the strength of an acid or base. The stronger the acid, the higher the value of Ka ; conversely, the stronger the base, the higher the value of Kb .

• Ka and Kb values help predict the pH of a solution.
• They are intrinsic properties of a substance.
• For instance, Kb for C6H5NH2 was used to calculate the ion concentrations in C6H5NH3NO3.

This knowledge is pivotal in pH calculations as it allows one to determine the concentration of H^+ or OH^- in weak acid or base solutions.

Neutralization Reactions

Neutralization reactions occur when an acid and a base react to form water and a salt. This process can either lead to a neutral, acidic, or basic solution, depending on the strengths of the reactants.
In the context of this exercise, understanding how neutralization affects pH is essential. A neutralization reaction between a strong acid and a strong base typically yields a neutral solution (pH=7). However, when weak acids or bases are involved, the result may be different.

• A solution of a salt from a strong acid and weak base will be acidic.
• A solution from a strong base and weak acid will be basic.

For example, KC7H5O2 results in a basic solution because it comes from KOH (a strong base) and HC7H5O2 (a weak acid). Understanding this concept helps in explaining the pH levels of salts derived from various acid-base combinations.