Calculate the pH of a buffer when 0.010 moles of NaOH is added to 100. mL solution that is 0.20 M sodium pentanoate (C4H9COONa) and 0.20 M pentanoic acid (C4H9COOH). pKa(C4H9COOH)

5.30

Explanation:

The pH of a buffer given by the Henderson-Hasselbach equation can be expressed as:

Given that:

Volume of the solution = 100.0 mL

To liters; the volume of the solution be 0.1 L

The concentration of = 0.20 M

The concentration of = molarity × volume

The concentration of = 0.20 mol/L × 0.1 L

The concentration of = 0.02 mol

The concentration of   = 0.20 M

The concentration of = 0.20 mol/L × 0.1 L

The concentration of = 0.02 mol

However, the number of moles of NaOH added = 0.01 moles

Now; The ICE table can be computed as:

                  C₄H₉COOH    +    OH⁻        ⇄     C₄H₉COO⁻      +       H₂O

Initial               0.02                  0.01                 0.02

Change          - 0.01                -0.01                 +0.01                     +0.01

Equilibrium      0.01                   -                      0.03                       0.01

Recall that the pH of = 4.82

∴

pH = 4.82 + log ( 3 )

pH = 4.82 + 0.4771

pH = 5.2971

pH ≅ 5.30

the best answer is d because the strong cohesion is caused by the ability of water molecules to undergo hydrogen bonding.

a is incorrect because when water solidifies, it becomes   ice, which expands and occupies a bigger volume for the same mass of water in its liquid state.

b is incorrect because water has high surface tension. besides bonding to itself it hydrogen bonds to other molecules at the top decreasing possibility for breaking of attractive forces.

c is incorrect because the water molecule has oxygen which is highly electronegative and so pulls the electron pair towards itself creating a dipole moment, making the molecule polar.

e is incorrect because water has strong adhesion, ability to stick to other molecules such as glass because of this molecules having stronger hydrogen bonding than water.