High School

Consider the titration of 46.5 mL of 1.2 M NaOH with 1.0 M HCl.

Find the pH at the following point:

- At the start of the titration.

Express your answer using two decimal places.

Answer :

At the start of the titration, there is no reaction between NaOH and HCl. Therefore, the pH can be calculated by determining the concentration of OH- ions in the NaOH solution. At the start of the titration, the pH is approximately 14.08. This can be calculated by determining the concentration of OH- ions in the NaOH solution and using the pOH equation to find the pH.


To calculate the pH, we can use the formula: pOH = -log[OH-]
Since NaOH is a strong base and fully dissociates in water, we can assume that the concentration of OH- ions is equal to the concentration of NaOH. First, let's calculate the number of moles of NaOH:
Moles of NaOH = Volume of NaOH solution (in L) x Concentration of NaOH (in mol/L)
Volume of NaOH solution = 46.5 mL = 0.0465 L
Concentration of NaOH = 1.2 M
Moles of NaOH = 0.0465 L x 1.2 mol/L = 0.0558 mol
Since NaOH is a strong base, it dissociates fully to produce 0.0558 mol of OH- ions.

Now, let's calculate the concentration of OH- ions: Concentration of OH- = Moles of OH- ions / Volume of solution (in L)
Since the volume of solution is the same as the volume of NaOH solution, the concentration of OH- ions is:
Concentration of OH- = 0.0558 mol / 0.0465 L = 1.2 M
Now, let's calculate the pOH:
pOH = -log[OH-] = -log(1.2) ≈ -0.08
To find the pH, we can use the equation:
pH + pOH = 14
pH = 14 - pOH = 14 - (-0.08) = 14.08
Therefore, at the start of the titration, the pH is approximately 14.08.

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