24.00 mL of a 0.25 M NaOH solution is titrated with 0.10M HCl. What is the pH of the solution after 24.00 mL of the HCl has been added

Question

Quizplus · Accepted Answer

Firstly, it is important to write a balanced chemical equation for the reaction that is occurring. In this case, it is:

NaOH + HCl → NaCl + H2O

From the equation, we can see that the reaction is an acid-base neutralization reaction, where NaOH is the base and HCl is the acid. This means that as the HCl is added, it will react with the NaOH to form NaCl and water.

To determine the pH at the end of the titration, we need to figure out how much HCl has reacted with the NaOH. We can do this using the following equation:

n(HCl) = c(HCl) x V(HCl)

where n is the number of moles of HCl, c is the concentration of HCl, and V is the volume of HCl added. Plugging in the values, we get:

n(HCl) = 0.10 mol/L x 0.02400 L = 0.00240 mol

Since the balanced equation tells us that the reaction between 1 mole of NaOH and 1 mole of HCl produces 1 mole of water, we know that the amount of NaOH that has reacted is also 0.00240 mol.

This means that there are now 0.0024 mol of NaOH remaining in the solution, since not all of it has reacted with the HCl. We can use this information to calculate the new concentration of NaOH:

c(NaOH) = n(NaOH) / V(total)

where V(total) is the total volume of the solution, which is equal to the sum of the volumes of NaOH and HCl. In this case, V(total) = 0.02400 L + 0.02400 L = 0.04800 L.

Plugging in the values, we get:

c(NaOH) = 0.00240 mol / 0.04800 L = 0.050 mol/L

Now that we know the concentration of NaOH, we can use the following equation to calculate the pH:

pH = 14 - log([OH-])

where [OH-] is the concentration of hydroxide ions, which is equal to the concentration of NaOH since NaOH is a strong base that dissociates completely in water.

Plugging in the value for the concentration of NaOH, we get:

pH = 14 - log(0.050 mol/L) = 12.88

Therefore, the final pH of the solution is 12.88. The correct answer is D.

2400 ML of a 0