A 50.0-mL sample of 0.10 M HNO2 is titrated with 0.10 M NaOH. What is the pH after 25.0 mL of NaOH

have been added?

To determine the pH after adding 25.0 mL of NaOH, we need to consider the reaction between HNO2 (nitrous acid) and NaOH (sodium hydroxide). Nitrous acid is a weak acid, and sodium hydroxide is a strong base. The reaction between them produces water (H2O) and a salt (NaNO2).

The balanced chemical equation for the reaction is:

HNO2 + NaOH -> NaNO2 + H2O

Since the reaction is a one-to-one ratio, 1 mole of HNO2 reacts with 1 mole of NaOH, yielding 1 mole of NaNO2 and 1 mole of H2O.

Before adding NaOH, we have a 50.0 mL sample of 0.10 M HNO2. To calculate the initial moles of HNO2, we use the formula:

moles = volume (L) x concentration (M)

moles of HNO2 = (50.0 mL / 1000 mL/L) x 0.10 M
= (0.0500 L) x 0.10 M
= 0.00500 moles HNO2

Since the reaction is a one-to-one ratio, the moles of HNO2 will be equal to the moles of NaOH when the reaction is complete. So, when 25.0 mL of NaOH (0.10 M) is added, the moles of NaOH reacted can be calculated as:

moles of NaOH = (25.0 mL / 1000 mL/L) x 0.10 M
= (0.0250 L) x 0.10 M
= 0.00250 moles NaOH

Now, we can calculate the remaining moles of HNO2:

remaining moles of HNO2 = initial moles of HNO2 - moles of NaOH reacted
= 0.00500 moles - 0.00250 moles
= 0.00250 moles HNO2

To find the concentration of HNO2 after the reaction, we divide the remaining moles by the total volume:

concentration of HNO2 = remaining moles of HNO2 / total volume (L)

The total volume is the sum of the initial volume of HNO2 and the volume of NaOH added:

total volume = (initial volume of HNO2 + volume of NaOH added) / 1000 (converted to L)

total volume = (50.0 mL + 25.0 mL) / 1000
= 75.0 mL / 1000
= 0.0750 L

Now we can calculate the concentration of HNO2 after the reaction:

concentration of HNO2 = 0.00250 moles / 0.0750 L
= 0.0333 M

Since we want to find the pH, we need to calculate the concentration of H+ ions. Nitrous acid is a weak acid, which partially dissociates in water. The dissociation equation for nitrous acid is:

HNO2 -> H+ + NO2-

The concentration of H+ ions is equal to the concentration of HNO2 after the reaction. Therefore, the concentration of H+ ions is 0.0333 M.

Finally, we can calculate the pH using the equation:

pH = -log[H+]

pH = -log(0.0333)
= 1.48 (rounded to two decimal places)

Therefore, the pH after adding 25.0 mL of NaOH is approximately 1.48.

Note: This calculation assumes that the final volume is equal to the sum of the initial volume of HNO2 and the volume of NaOH added. It also assumes that there are no other acid-base reactions or significant ionizations occurring.