Calculate the molarity of each ion in 0.300 M Na2SO4. Enter a zero for an ion that doesn't occur in the compound.
______ M Na1+
______ 2 M S2-
______ 3 M O 2-
______ M SO42-
0.3M Na2SO4 means you have
0.3 moles Na2SO4/L soln.
So you will have 0.3M Na2SO4, 0.6M Na^+, 0.3M S, 4*0.3M O and 0.3M SO4^2-.
I have no idea what 2 M S2- or 3 M O 2- is.
To calculate the molarity of each ion in Na2SO4, we need to understand the formula of the compound. Na2SO4 consists of two sodium ions (Na1+), one sulfate ion (SO42-), and no additional oxygen ions (O2-).
To determine the molarity of each ion, we can refer to the coefficients in the formula. The coefficients indicate the number of each ion present in one formula unit of the compound.
1. Na2SO4: The coefficient of Na1+ is 2, which means there are 2 Na1+ ions in one formula unit of Na2SO4. Therefore, the molarity of Na1+ is also 2 times the overall molarity of Na2SO4.
Molarity of Na1+ = 2 * 0.300 M = 0.600 M
2. S2-: The coefficient of S2- in Na2SO4 is 1. Hence, the molarity of S2- is equal to the overall molarity of Na2SO4.
Molarity of S2- = 0.300 M
3. O2-: There is no separate O2- ion present in Na2SO4. Hence, the molarity of O2- is zero.
Molarity of O2- = 0 M
4. SO42-: The coefficient of SO42- is also 1 in Na2SO4. Thus, the molarity of SO42- is the same as the overall molarity of Na2SO4.
Molarity of SO42- = 0.300 M
So, the molarity of each ion in 0.300 M Na2SO4 is as follows:
M Na1+ = 0.600 M
M S2- = 0.300 M
M O2- = 0 M
M SO42- = 0.300 M