1990 Iowa Senate election

The 1990 Iowa State Senate elections took place as part of the biennial 1990 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 1990 is provided by the Iowa General Assembly here.

The primary election on June 5, 1990, determined which candidates appeared on the November 6, 1990 general election ballot. Primary election results can be obtained here.^[1] General election results can be obtained here.^[2]

The 1990 elections were the last in Iowa in which the Lieutenant Governor performed the duties of "President of the Senate." Starting on January 14, 1991, with the enactment of Article IV, section 18, of the Constitution of Iowa, the duties of Iowa's Lieutenant Governor no longer include presiding over the state Senate.^[3] The Majority Leader was instead the sitting Senate member who led the larger party. Following 1991, the President of the Iowa Senate would become a sitting member of the Senate.

Following the previous election in 1988, 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 kept their control of the Iowa State Senate following the 1990 general election, with Democrats holding 28 seats and Republicans having 22 seats after the election (a net gain of 2 seats for the Republicans).