A500 ml sample of a 0.100 m formate buffer, ph 3.75, is treated with 5 ml of 1.00 m koh. what is the ph following this addition?

The pH of the solution after addition of KOH is 3.84

Explanation:

We are given:

pH of buffer = 3.75

of formic acid = 3.75

Using Henderson-Hasselbalch equation for formate buffer:

Putting values in above equation, we get:

We are given:

Concentration of formate buffer = 0.100 M

As, the volume of buffer is the same. So, the concentration is taken as number of moles of formate ions as well as formic acid

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

Molarity of KOH = 1.00 M

Volume of solution = 5 mL

Putting values in above equation, we get:

The chemical reaction for formic acid and KOH follows the equation:

Initial:       0.05    0.005               0.05

Final:         0.045          -                0.055          

Volume of solution = 500 + 5 = 505 mL = 0.505 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 formic acid = 3.75

pH = ?

Putting values in above equation, we get:

Hence, the pH of the solution after addition of KOH is 3.84

a graduated cylinder.

explanation:

answer: c. the experiment shows that the red substance experienced a chemical change.

explanation: chemical change is the change in which chemical composition of the substance changes. physical change is the change in which there is no change in chemical composition.

chemical changes are accompanied by following changes:

a) precipitate formation

b) energy is released or absorbed

c) formation of bubbles

d) change in color

as it is given that the red color has disappeared and there is production of gas, it means the red substance has undergone a chemical change.

e