A 50 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M HCl. What is the initial pH?

The question is –

A 50 mL sample of 0.200 M sodium hydroxide is titrated with 0.200 M HCl. What is the initial pH?

Answer:

⇒ Initial pH is 13.301.

Explanation:

First, we need to understand what we’re dealing with here. Sodium hydroxide, NaOH, is a strong base, and hydrochloric acid, HCl, is a strong acid. When in solution, they dissociate completely.

For NaOH, we can write this dissociation as follows:

NaOH_(aq) → Na⁺_(aq) + OH^–_(aq)

Now, let’s calculate the initial concentration of OH- ions in the solution.

The molarity of NaOH in the solution is given as 0.200 M, and because each molecule of NaOH provides one OH- ion, the initial concentration of OH- ions in the solution is also 0.200 M.

Next, let’s use this concentration to calculate the initial pOH of the solution. The pOH is defined as the negative logarithm (base 10) of the concentration of OH- ions.

⇒ pOH = -log[OH-]

Plug in the concentration of OH- ions:

∴ pOH = -log(0.200)

Let’s calculate this:

∴ pOH = 0.699 (rounded to three decimal places)

Since the sum of pH and pOH is equal to 14 at 25 degrees Celsius:

⇒ pH + pOH = 14

We can rearrange this to solve for pH:

⇒ pH = 14 – pOH

Substitute in the pOH we just calculated:

⇒ pH = 14 – 0.699

Calculate this:

∴ pH = 13.301

So, the initial pH of the 0.200 M NaOH solution is approximately 13.3 (rounded to one decimal place).

Note: These calculations assume that the temperature is 25 degrees Celsius, because the relationship between pH, pOH, and 14 is temperature-dependent.