Question 8: a 150 ml of buffer contains 0.045 m carbonic acid (h2co3) and 0.025 m sodium hydrogen carbonate (hco3). if 3.0 ml of 0.15m hcl is added to the buffer, what is the new ph? pka = 6.37

The new pH of resulting solution is 6.03

Explanation:

We are adding hydrochloric acid to the solution, so it will react with salt (sodium hydrogen carbonate) only.

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

Molarity of hydrochloric acid = 0.15 M

Volume of solution = 3 mL

Putting values in above equation, we get:

Molarity of sodium hydrogen carbonate = 0.025 M

Volume of solution = 150 mL

Putting values in above equation, we get:

Molarity of carbonic acid = 0.045 M

Volume of solution = 150 mL

Putting values in above equation, we get:

The chemical reaction for sodium hydrogen carbonate and hydrochloric acid follows the equation:

Initial:        0.00375    0.00045                0.00675

Final:         0.00330          -                       0.00720          

Volume of solution = 150 + 3 = 153 mL = 0.153 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 carbonic acid = 6.37

pH = ?

Putting values in above equation, we get:

Hence, the new pH of the solution is 6.03

lol idk man i guess u gotta ask bill nye the science guy

6.4 cm (max) - 25 cm (min)