1988 Iowa Senate election

1988 Iowa Senate election

← 1986 November 8, 1988 1990 →

25 out of 50 seats in the Iowa State Senate
26 seats needed for a majority
  Majority party Minority party
 
Leader Bill Hutchins Calvin Hultman
Party Democratic Republican
Leader's seat 48th 47th
Last election 30 20
Seats before 30 20
Seats after 30 20
Seat change Steady Steady

Majority Leader before election

Bill Hutchins
Democratic

Elected Majority Leader

Bill Hutchins
Democratic

The 1988 Iowa State Senate elections took place as part of the biennial 1988 United States elections. Iowa voters elected state senators in half of the state senate's districts—the 25 even-numbered state senate districts. State senators serve four-year terms in the Iowa State Senate, with half of the seats up for election each cycle. A statewide map of the 50 state Senate districts in the year 1988 is provided by the Iowa General Assembly here.

The primary election on June 7, 1988 determined which candidates appeared on the November 8, 1988 general election ballot. Primary election results can be obtained here.[1] General election results can be obtained here.[2]

Following the previous election, Democrats had control of the Iowa state Senate with 30 seats to Republicans' 20 seats.

To take control of the chamber from Democrats, the Republicans needed to net 6 Senate seats.

Democrats retained control of the Iowa State Senate following the 1988 general election with the balance of power unchanged: Democrats holding 30 seats and Republicans having 20 seats after the 1988 election.

  1. ^ "Primary Election 1988 Canvass Summary" (PDF). Iowa Secretary of State. Retrieved April 17, 2020.
  2. ^ "General Election 1988 Canvass Summary" (PDF). Iowa Secretary of State. Retrieved April 17, 2020.

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