{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ch302 notes ch7.2

# ch302 notes ch7.2 - Example In a 0.12 M solution of a weak...

This preview shows pages 1–2. Sign up to view the full content.

CH302 Chapter 7 notes part 2 ACIDS and BASES pH of a WEAK Acid In solution we have two potential acids: HA (aq) H + (aq) + A - (aq) H 2 O (l) H + (aq) + OH - (aq) BOTH species can provide H + but K w is often MUCH smaller than K a so we assume HA is the strongest acid. [ ][ ] [] HA A O H K 3 a + = Example: Finding pH for a Weak Monoprotic Acid Calculate the concentrations of the various species in 0.15 M acetic acid, CH 3 COOH, solution. Percent Dissociation Percent Dissociation = amount dissociated(mol/L) x 100% initial concentration(mol/L) Another way to see how strong an acid is. A STRONG acid will be 100% dissociated.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Example: In a 0.12 M solution of a weak monoprotic acid, HY, the acid is 5.0% ionized. Calculate K a for this acid. Using Ka to get pH, the % dissociation: The pH of a 0.10 M solution of a weak monoprotic acid, HA, is found to be 2.97. What is the value for K a ? Now calculate the percent ionization for the 0.15 M acetic acid. Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. e'll fill this out and compare them: K a = 4.0 x 10-10 for HCN W 0.15 M HCN 0.15 M acetic acid % pH [H + ] K a Solution...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

ch302 notes ch7.2 - Example In a 0.12 M solution of a weak...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online