PHP - sacar en una variable un dato de un array

$ip_address = ['192.168.1.1', '192.168.1.2'];

$first_ip = explode('.', $ip_address[0]);

echo $first_ip[0] . '.' . $first_ip[1] ; //192.168

$ContadorVisitasHoy ="SELECT distinct ip FROM contador2 where Fecha ='$FechaHoy'";

			$SqlResult  = mysqli_query( $conexion, $ContadorVisitasHoy ) or die ( "Algo ha ido mal en la consulta a la base de datos");

			while ($row= mysqli_fetch_array($SqlResult, MYSQLI_ASSOC))

				{

					$first_ip = explode('.', $row[0]);

					echo $first_ip[0] . '.' . $first_ip[1] ; //192.168

				}

PHP Warning:  explode() expects parameter 2 to be string, array given in C:\inetpub\wwwroot\admin2.php on line 123

PHP Notice:  Trying to access array offset on value of type null in C:\inetpub\wwwroot\admin2.php on line 124

$first_ip = explode('.', $row['ip']);