{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

FINAL_REVIEW_part_2

# FINAL_REVIEW_part_2 - 1.Titration of weak acid with a...

This preview shows pages 1–5. Sign up to view the full content.

1. Titration of weak acid with a strong base: a. Titrate 15mL of 1.2M HF (Ka = 7.2 × 10 -4 ) with a 2.0M NaOH solution. Find pH: When no base added 4.5mL of base added 6.25mL of base added 9mL of base added 13mL of base added General equation of weak acid w/ strong base: HA + OH - H 2 O + A - Equation for this problem: HF + NaOH NaF + H 2 O No Base Added—Point A: - What is in Solution?—only weak acid (HA), in this case HF - Since [HA] i ≠ [H 3 O + ], need to do a Ka ICE table equilibrium calculation to find [H 3 O + ].

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

View Full Document
- - 4.5mL of Base Added—Region A': - Has some OH - - OH - reacts immediately to form some H 2 O and A - - HA + H 2 O H 3 O + + A - Stress: [HA]↓, [A - ]↑ Equilibrium shifts left, suppresses formation of [H 3 O + ] pH takes an initial “jump” 6.25mL Base Added—Region B: - More OH - added causes more A - formation - Initial mol HA > mol OH - added - What is in solution here?—some A - and some unreacted HA - This is a BUFFER! b/c weak acid + conj. base—use

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.

{[ snackBarMessage ]}

### Page1 / 5

FINAL_REVIEW_part_2 - 1.Titration of weak acid with a...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online