Broward County, FL vs Snohomish County, WA
Property Tax Rate Comparison 2025
Quick Answer
Broward County, FL: 0.88% effective rate · $3,539/yr median tax · median home $402,176
Snohomish County, WA: 0.79% effective rate · $3,061/yr median tax · median home $387,523
Side-by-Side Comparison
| Metric | Broward County, FL | Snohomish County, WA | National Avg |
|---|---|---|---|
| Effective Tax Rate | 0.88% | 0.79% | 1.06% |
| Median Annual Tax | $3,539 | $3,061 | $2,778 |
| Median Home Value | $402,176 | $387,523 | $268,728 |
| Population | 1,944,375 | 827,957 | — |
Broward County, FL
- Effective Rate
- 0.88%
- Median Annual Tax
- $3,539
- Median Home Value
- $402,176
- Population
- 1,944,375
Snohomish County, WA
- Effective Rate
- 0.79%
- Median Annual Tax
- $3,061
- Median Home Value
- $387,523
- Population
- 827,957
Frequently Asked Questions
What is the property tax difference between Broward County and Snohomish County?
Broward County, FL has an effective property tax rate of 0.88% with a median annual tax of $3,539. Snohomish County, WA has a rate of 0.79% with a median annual tax of $3,061. The difference is 0.09 percentage points.
Which county has higher property taxes, Broward County or Snohomish County?
Broward County, FL has the higher effective property tax rate at 0.88% compared to 0.79%.
How do Broward County and Snohomish County compare to the national average?
The national average effective property tax rate is 1.06%. Broward County is below average at 0.88%, and Snohomish County is below average at 0.79%.