Calculate the pH of adding 20 mL of 1 M NaOH solution to 100 mL of a 1 M acetic acid (CH3COOH) solution and 880 mL of distilled water. Assume that the Ka of acetic acid is 1.8 * 10-5

The pH of the solution is 4.14

Explanation:

To calculate the number of moles for given molarity, we use the equation:

Molarity of NaOH = 1 M

Volume of solution = 20 mL

Putting values in above equation, we get:

Molarity of acetic acid solution = 0.26 M

Volume of solution = 100 mL

Putting values in above equation, we get:

The chemical reaction for NaOH and acetic acid follows the equation:

Initial:           0.100            0.020

Final:            0.080               -              0.020            

Volume of solution = 20 + 100 = 120 mL = 0.120 L    (Conversion factor:  1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

We are given:

= negative logarithm of acid dissociation constant of acetic acid = 4.74

pH = ?

Putting values in above equation, we get:

Hence, the pH of the solution is 4.14

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .

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