1.50 g of a weak acid (molar mass 176) is dissolved in 50.0 ml of water, and the resultant solution is titrated with 0.300 m naoh. when 12.5 ml of 0.300 m naoh is added, the ph of the resultant solution is 4.00. calculate ka for this acid.

The dissociation constant of this acid is .

Explanation:

Moles of acid =

Moles of sodium hydroxide:

Moles of sodium hydroxide:

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

Initial:        0.008523    0.00375                0

Final:       (0.008523-0.00375)   0              0.00375

Volume of solution = 50.0 mL + 12.5 mL = 62.5 mL = 0.0625 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:

= ?

pH = 4.00

Putting values in above equation, we get:

The dissociation constant of this acid is .

explanation:

0.07 g/cm³

1.37 g/cm³

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.