Question

Calculate the pH of the solution that results following addition of 25 mL of 1.0 MNaOH to 45 mL of 1.0 MNH₄Cl.

Express your answer to two decimal places.

Answer 1

To calculate the pH of the solution resulting from the addition of NaOH to NH₄Cl, we need to determine the moles of NH₄Cl and NaOH and use them to calculate the concentration of the resulting NH₄OH solution. Then, we can use the concentration of NH₄OH to calculate the pOH and finally convert it to pH. The pH of the solution is 13.81.

Step-by-step explanation:

To calculate the pH of the solution resulting from the addition of NaOH to NH₄Cl, we need to determine the moles of NH₄Cl and NaOH and use them to calculate the concentration of the resulting NH₄OH solution. Then, we can use the concentration of NH₄OH to calculate the pOH and finally convert it to pH.

First, let's calculate the moles of NH₄Cl:

Mole of NH₄Cl = Volume (L) × Concentration (M) = 45 mL × 0.001 M/mL = 0.045 M

Next, let's calculate the moles of NaOH:

Mole of NaOH = Volume (L) × Concentration (M) = 25 mL × 0.001 M/mL = 0.025 M

Since the reaction between NH₄Cl and NaOH is a 1:1 ratio, the moles of NH₄OH formed will be equal to the moles of NH₄Cl used:

Mole of NH₄OH = 0.045 M

Now, let's calculate the concentration of NH₄OH in the solution:

Volume of NH₄OH = Volume of NH₄Cl + Volume of NaOH = 45 mL + 25 mL = 70 mL = 0.07 L

Concentration of NH₄OH = Mole of NH₄OH / Volume (L) = 0.045 M / 0.07 L = 0.643 M

Now, we can calculate the pOH using the concentration of NH₄OH:

pOH = -log[OH-] = -log(0.643) = 0.19

Finally, we can convert the pOH to pH using the equation: pH = 14 - pOH:

pH = 14 - 0.19 = 13.81