Wakulla County Fire Rescue Firefighter Salary
Crawfordville, Florida · Wakulla County · 33 career firefighters · 11 stations
What Wakulla County Fire Rescue pays
Entry-level starting pay
$53,235 to $54,109
a year
Top of posted range
Not listed
a year
Entry-level postings
2 postings
May 2026 to May 2026
Wakulla County Fire Rescue listed starting pay of $53,235 to $54,109 a year in 2 entry-level postings between May 2026 and May 2026. The most recent, Fire Fighter/Paramedic in May 2026, started at $54,109.
2 other postings quoted hourly pay of $15.61 to $17.59 an hour.
Past Wakulla County Fire Rescue postings with pay
Closed postings FireJobs recorded in the last two years. Figures are as the department listed them.
| Position | Posted | Starting | Top of range |
|---|---|---|---|
| Fire Fighter/Paramedic | May 2026 | $54,109 | Not listed |
| Paramedic | May 2026 | $53,235 | Not listed |
Wakulla County Fire Rescue pay questions
- How much does a firefighter at Wakulla County Fire Rescue make?
- Wakulla County Fire Rescue listed starting pay of $53,235 to $54,109 a year in 2 entry-level postings between May 2026 and May 2026. The most recent, Fire Fighter/Paramedic in May 2026, started at $54,109.
- How does Wakulla County Fire Rescue pay compare with other Florida departments?
- The average starting salary across 52 current Florida firefighter listings on FireJobs is $65,234. Wakulla County Fire Rescue's most recent starting figure of $54,109 is 17% below it.
- Is Wakulla County Fire Rescue hiring right now?
- Yes. Wakulla County Fire Rescue has 3 open positions on FireJobs that list pay. Members see the salary, closing date and the application link.
Source: pay figures the department published in its own job postings, recorded by FireJobs when each posting was live. Pay changes with contracts and steps; the posting for a current opening is the figure that applies.