## pH Calculation

## What is pH

pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+) in a solution. The pH scale ranges from 0 to 14, where a pH of 7 is neutral (indicating that the concentration of H+ ions and hydroxide ions, OH-, are equal), a pH less than 7 is acidic (indicating a higher concentration of H+ ions), and a pH greater than 7 is basic (indicating a higher concentration of OH- ions).

## How pH is calculated

The pH of a solution can be calculated using the formula:

pH = -log[H+]

where [H+] is the concentration of hydrogen ions in moles per liter (M) in the solution.

For example, if the concentration of H+ ions in a solution is 1 x 10^-5 M, the pH can be calculated as follows:

pH = -log(1 x 10^-5) pH = 5

So, the pH of the solution is 5.

In practice, it is often more convenient to use a pH meter or pH indicator paper to measure the pH of a solution rather than to calculate it using the formula above. The pH meter measures the electrical potential difference between a pH electrode and a reference electrode in the solution, which is converted into a pH value. pH indicator paper contains a pH-sensitive dye that changes color in response to changes in pH, which can be compared to a color chart to determine the pH of the solution.

## Question 1:

A buffer solution consists of 1 M CH3COOH and 1 CH3COONa. The

ionization constant of the acid is 1.8 x 10-5. Calculate the pH:

a) Of that buffer solution consisting of 1 M CH3COOH and 1 M

CH3COONa

b) After adding 0.040 mol of solid NaOH to 500 mL of the buffer solution

(assume the addition of NaOH cause a negligible change in volume)

c) After adding 0.040 mol of HCl to 500 mL of the buffer solution

(assume the addition of HCl cause a negligible change in volume)

## Answer

a) To calculate the pH of the buffer solution, we need to first determine the pKa of CH3COOH, which can be calculated using the ionization constant as follows:

pKa = -log(Ka) = -log(1.8 x 10^-5) = 4.74

The pH of the buffer solution can then be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([CH3COO-]/[CH3COOH])

where [CH3COO-] is the concentration of the acetate ion and [CH3COOH] is the concentration of the acetic acid. In this case, both concentrations are 1 M, so we have:

pH = 4.74 + log(1/1) = 4.74

Therefore, the pH of the buffer solution is 4.74.

b) When 0.040 mol of solid NaOH is added to the buffer solution, it reacts with the acetate ion to form more CH3COOH and water, according to the following balanced chemical equation:

CH3COO- + NaOH → CH3COOH + Na+ + OH-

The amount of acetate ion is initially 1 M, but some of it reacts with the added NaOH. The amount of acetic acid also increases as a result of the reaction, but since the volume of the solution is assumed to be constant, the total buffer concentration remains 2 M. To calculate the new pH of the buffer solution, we need to determine the concentrations of CH3COOH and CH3COO- after the reaction has occurred.

Since the added NaOH is 0.040 mol, it will completely react with 0.040 mol of CH3COO-. The remaining acetate ion concentration is then:

[CH3COO-] = 1 M – (0.040 mol / 0.5 L) = 0.92 M

The concentration of acetic acid can be calculated using the buffer equation and the new acetate ion concentration:

pH = pKa + log([CH3COO-]/[CH3COOH]) 4.74 = 4.74 + log(0.92/[CH3COOH]) [CH3COOH] = 0.92/10^(4.74-4.74) = 0.92 M

Therefore, the new concentration of CH3COOH is 0.92 M. Using the Henderson-Hasselbalch equation, the new pH of the buffer solution can be calculated:

pH = pKa + log([CH3COO-]/[CH3COOH]) pH = 4.74 + log(0.92/0.92) = 4.74

The new pH of the buffer solution is still 4.74, since the buffer is able to resist changes in pH when small amounts of acid or base are added.

c) When 0.040 mol of HCl is added to the buffer solution, it reacts with the acetate ion to form more acetic acid and chloride ion, according to the following balanced chemical equation:

CH3COO- + HCl → CH3COOH + Cl-

The amount of acetate ion is initially 1 M, but some of it reacts with the added HCl. The amount of acetic acid also increases as a result of the reaction, but since the volume of the solution is assumed to be constant, the total buffer concentration remains 2 M. To calculate the new pH of the buffer solution, we need to determine the concentrations of CH3COOH and CH3COO- after the reaction has occurred.

Since

the added HCl is 0.040 mol, it will completely react with 0.040 mol of CH3COO-. The remaining acetate ion concentration is then:

[CH3COO-] = 1 M – (0.040 mol / 0.5 L) = 0.92 M

The concentration of acetic acid can be calculated using the buffer equation and the new acetate ion concentration:

pH = pKa + log([CH3COO-]/[CH3COOH]) 4.74 = 4.74 + log(0.92/[CH3COOH]) [CH3COOH] = 0.92/10^(4.74-4.74) = 0.92 M

Therefore, the new concentration of CH3COOH is 0.92 M. Using the Henderson-Hasselbalch equation, the new pH of the buffer solution can be calculated:

pH = pKa + log([CH3COO-]/[CH3COOH]) pH = 4.74 + log(0.92/0.92) = 4.74

The new pH of the buffer solution is still 4.74, since the buffer is able to resist changes in pH when small amounts of acid or base are added.

In conclusion, the pH of the buffer solution remains unchanged after the addition of either 0.040 mol of solid NaOH or 0.040 mol of HCl, since the buffer is able to resist changes in pH when small amounts of acid or base are added.

## Question 2:

Calculate the pH of a solution that is 0.0480 M in

a. Na2Cr2O4 (Ka2 = 5.42*10-5) pH = ______

b. Na2SO3 (Ka2 = 6.6*10-8) pH = ______

c. C2H4 (NH3+Cl-)2 (Ka1 = 1.42*10-7) pH = ______

## Answer

a. Na2Cr2O4 dissociates into Na+ and the polyatomic ion Cr2O42-, which can undergo hydrolysis to form HCr2O4- and OH-. The equilibrium constant for this reaction is Ka2 = [HCr2O4-][OH-]/[Cr2O42-]. Since Na2Cr2O4 is a strong electrolyte, we can assume that all of the Cr2O42- ions dissociate. Therefore, [Cr2O42-] = 0.0480 M.

Let x be the concentration of OH-. Then [HCr2O4-] = [OH-] = x, and [Cr2O42-] = 0.0480 – x. Substituting these into the equilibrium constant expression, we get:

Ka2 = x^2 / (0.0480 – x)

Solving for x using the quadratic formula, we get:

x = 4.70 × 10^-6 M

Since [H+] = Kw / [OH-], where Kw is the ion product constant for water (1.0 × 10^-14 at 25°C), we can calculate the pH as:

pH = -log[H+] = 8.32

b. Na2SO3 dissociates into Na+ and the polyatomic ion SO32-, which can undergo hydrolysis to form HSO3- and OH-. The equilibrium constant for this reaction is Ka2 = [HSO3-][OH-]/[SO32-]. Since Na2SO3 is a strong electrolyte, we can assume that all of the SO32- ions dissociate. Therefore, [SO32-] = 0.0480 M.

Let x be the concentration of OH-. Then [HSO3-] = [OH-] = x, and [SO32-] = 0.0480 – x. Substituting these into the equilibrium constant expression, we get:

Ka2 = x^2 / (0.0480 – x)

Since Ka2 is very small compared to the initial concentration of SO32-, we can assume that x << 0.0480. Under this assumption, we can simplify the equilibrium constant expression to:

Ka2 = x^2 / 0.0480

Solving for x, we get:

x = 2.07 × 10^-5 M

Therefore, the pH of the solution is:

pH = -log[H+] = 9.68

c. C2H4(NH3+Cl-)2 is a weak acid with a dissociation constant of Ka1 = [H+][C2H4(NH3+Cl-)]/[C2H4(NH3+Cl-)2]. Let x be the concentration of H+, which is equal to the concentration of C2H4(NH3+Cl-), since they dissociate in a 1:1 ratio. Then [C2H4(NH3+Cl-)2] = 0.0480 – x.

Substituting these into the equilibrium constant expression, we get:

Ka1 = x^2 / (0.0480 – x)

Solving for x using the quadratic formula, we get:

x = 2.55 × 10^-4 M

Therefore, the pH of the solution is:

pH = -log[H+] = 3.59