1998 Iowa Senate election

← 1996 November 3, 1998 2000 →
25 out of 50 seats in the Iowa State Senate 26 seats needed for a majority
Majority party Minority party   Leader Mary Kramer Michael Gronstal Party Republican Democratic Leader's seat 37th 42nd Last election 29 21 Seats before 28 22[a] Seats after 30 20 Seat change 2 2
President of the Senate before election Mary Kramer Republican Elected President of the Senate Mary Kramer Republican

Elections in Iowa
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v t e

The 1998 Iowa State Senate elections took place as part of the biennial 1998 United States elections. Iowa voters elected state senators in half of the state senate's districts—the 25 odd-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 1998 is provided by the Iowa General Assembly here.

The primary election on June 2, 1998 determined which candidates appeared on the November 3, 1998 general election ballot. Primary election results can be obtained here.^[2] General election results can be obtained here.^[3]

Following the previous election in 1996, Republicans had control of the Iowa state Senate with 29 seats to Democrats' 21 seats. In a special election in District 13 in 1997 to fill a vacancy created by Sen. Jim Lind's (R) resignation, Democratic candidate Patricia M. "Pat" Harper (D) flipped the seat for her party. Therefore, on election day 1998, Republicans held 28 seats to Democrats' 22.

To reclaim control of the chamber from Republicans, the Democrats needed to net 4 Senate seats.

Republicans strengthened their control of the Iowa State Senate following the 1998 general election by netting two seats, resulting in Republicans holding 30 seats and Democrats having 20 seats after the election.

Summary of Results

• NOTE: The 25 even-numbered districts did not have elections in 1998 so they are not listed here.