Find the pH of this buffer after 0.069 mol of OH- was added to 1.00 l of this buffer containing 0.165 M is and 0.594 (Ka₁ = 7.2 x 10⁻³; Ka₂ = 6.1 x 10⁻⁸; Ka₃ = 4.2 x 10⁻¹³) assume no volume change due to the addition of OH⁻.

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

both adhesion and cohesion are attractions.

they are different in that adhesion is the force of attraction between the molecules of two different substances, whereas cohesion is the force of attraction between molecules of the same substance.

possible examples include capillary action, in which the adhesion between water and glass causes the water to rise in a narrow tube against the force of gravity.

because of hydrogen bonding, water is extremely cohesive.

this causes molecules on the surface of water to be drawn inward, which is why drops of water form beads on smooth surfaces.

hope this .