Question 4 (1 point) 60.5 mL of 0.223 M HI is mixed with 86.0 mL of a buffer system composed of 0.385 M sodium acetate and 0.410 M acetic acid. Find the pH of

the final solution

Ko of acetic acid = 1.8 E-5

4561

o a

Ob

Ос

Od

4.673

4.450

4.349

pH = 4.349

Explanation:

When HI is added to sodium acetate, CH3COONa, acetic acid is produced as follows:

CH3COONa + HI → CH3COOH + NaI

That means the moles of HI added are the moles of acetic acid produced and the moles of sodium acetate consumed.

Moles of HI:

0.0605L * (0.223mol/L) = 0.01349mol

To find pH of the buffer (Mixture of weak acid, acetic acid, and conjugate base, sodium acetate) we need to solve H-H euation:

pH = pKa + log [CH3COONa] / [CH3COOH]

Where pKa is -log of Ka of the acid: -log1.8x10⁻⁵ = 4.745

And [] could be taken as moles of each specie

Moles CH3COONa:

Inital moles: 0.086L * (0.385mol/L) = 0.03311

Final moles: 0.03311mol - 0.01349mol = 0.01962mol CH3COONa

Moles CH3COOH:

Inital moles: 0.086L * (0.410mol/L) = 0.03256

Final moles: 0.03256mol + 0.01349mol = 0.04875mol CH3COOH

pH = 4.745+ log [0.01962mol CH3COONa] / [0.04875mol CH3COOH]

pH = 4.3497

Rigth answer is:

answer; /// i do believe that the correct answer is (a) boiling

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.