Calculate the ph of 0.375 l of a 0.18 m acetic acid-0.29 m sodium acetate buffer after the addition of 0.0090 mol of koh. assume that the volume remains constant.

To calculate the pH of this solution, we use the Henderson-Hasselbalch equation: 

pH = pKa + log ([A-]/[HA])

Where,

[A-] = Molarity of the conjugate base = CH3COO- = 0.29 M
[HA] = Molarity of the weak acid  = CH3COOH = 0.18 M

pKa = dissociation constant of the weak acid = 4.75

When KOH is added to the buffer, the chemical reaction is:

CH3COOH + KOH = CH3COO-K+ + H2O

Therefore when 0.0090 mol KOH is added, 0.0090 mol acid is neutralized, and 0.0090 mol CH3COO- is produced.

[CH3COO-] = [0.0090 mol + 0.375 L (0.29 mol/L) ] / 0.375 L = 0.314 M

[CH3COOH] = [-0.0090 mol + 0.375 L (0.18 mol/L) ] / 0.375 L = 0.156 M

Going back to Henderson-Hasselbalch equation:

pH = 4.75 + log (0.314 / 0.156)

pH = 5.054

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

m=9

step-by-step explanation: