Why is a whitespace character only represented by 6 bits in ASCII?

I have written a code in python to represent strings using their ASCII counterparts. I have noticed that every character is replaced by 7 bits (as I expected). The problem is that every time I include a space in the string I am converting it is only represented by 6 bits instead of 7. This is a bit of a problem for a Vernam Cipher program I am writing where my ASCII code is always a few bits smaller than my key due to spaces. Here is the code and output below:

string = 'Hello t'
ASCII = ""
for c in string:
    ASCII += bin(ord(c)) 
ASCII = ASCII.replace('0b', ' ')

print(ASCII)

Output: 1001000 1100101 1101100 1101100 1101111 100000 1110100

As can be seen in the output the 6th sequence of bits which represents the space character has only 6 bits and not 7 like the rest of the characters.

Instead of bin(ord(c)), which will automatically strip leading bits, use string formatting to ensure a minimum width:

f'{ord(c):07b}'

The problem lies within your "conversion" - the value for whitespace happens to only need 6 bits, and the bin built-in simply don't do left padding with zeros. That is why you are getting 7 bits for other chars - but it would really be more confortable if you would use 8 bits for everything.

One way to go is, instead of using the bin call, use string formatting operators: these can, besides the base conversion, pad the missing bits with 0s:

string = 'Hello t'
# agregating values in a list so that you can easily separate the binary strings with a " "
# by using ".join"
bin_strings = []
for code in string.encode("ASCII"): # you really should do this from bytes -
     #which are encoded text. Moreover, iterating a bytes
     # object yield 0-255 ints, no need to call "ord"
     bin_strings.append(f"{code:08b}")  # format the number in `code` in base 2 (b), with 8 digits, padding with 0s 
ASCII = ' '.join(bin_strings)

Or, as a oneliner:

ASCII = " ".join(f"{code:08b}" for code in "Hello World".encode("ASCII"))