The Pope’s apostolic trip to the USA, including a visit to the UN headquarters, will take place from 15th to 21st April. It has been officially confirmed today by the newsroom of the Holy See, which announced the schedule of Benedict XVI’s eighth apostolic journey outside Italy. The Pope’s plane will leave on 15th April at 12 from the airport of Fiumicino to land at 4 pm at the Air Force Base airport in Washington, where the Pope will be received (privately) by president Bush and the first lady. Then the Pope will go to the apostolic nunciature in Washington, where the morning after (16th April) he will say Mass in private before going to the White House for the welcome ceremony (10.30 am), with the Holy Father’s speech and a courtesy visit to the US president in the Oval Room. At 1 pm, he will lunch with the US cardinals with the "preasidium" of the American Catholic Bishops Conference and the Pope’s retinue in the apostolic nunciature in Washington, where at 04.45 pm the delegates of Catholic charities of the apostolic nunciature of Washington will come to greet him. At 5 pm, the Pope will be driven to the National Shrine of the Immaculate Conception in Washington, then at 05.45 pm he will celebrate the Vespers 17.45, then he will meet the US bishops and then will give a speech. (continued)
