what is the total molarity of all the ions present in 0.1M cuso4 and 0.1M al2(so4)3 solution?
Molarity=0.1+0.1+2×0.1+3×0.1
=0.1+0.1+0.2+0.3
=0.7M
To find the total molarity of all the ions present in the solution, we need to calculate the molarity of each ion separately and then add them up.
First, let's start with CuSO4:
- CuSO4 dissociates into one Cu2+ ion and one SO42- ion.
The molarity of Cu2+ ion in the CuSO4 solution is 0.1 M.
The molarity of SO42- ion in the CuSO4 solution is 0.1 M.
Next, let's move on to Al2(SO4)3:
- Al2(SO4)3 dissociates into two Al3+ ions and three SO42- ions.
The molarity of Al3+ ions in the Al2(SO4)3 solution is also 0.1 M (as it contains two Al3+ ions).
The molarity of SO42- ions in the Al2(SO4)3 solution is 0.1 M * 3 = 0.3 M (as it contains three SO42- ions).
Finally, let's sum up the molarities of all the ions:
- The total molarity of Cu2+ ions is 0.1 M.
- The total molarity of Al3+ ions is 0.1 M + 0.1 M = 0.2 M.
- The total molarity of SO42- ions is 0.1 M + 0.3 M = 0.4 M.
Therefore, the total molarity of all the ions present in the solution is 0.1 M Cu2+ + 0.2 M Al3+ + 0.4 M SO42- = 0.7 M.
To find the total molarity of all the ions present in the solution, we first need to determine the number of ions produced by each compound when it dissociates in water.
Copper sulfate (CuSO4) dissociates into one copper ion (Cu2+) and one sulfate ion (SO4^2-) when dissolved in water.
Aluminum sulfate (Al2(SO4)3) dissociates into two aluminum ions (Al^3+) and three sulfate ions (SO4^2-) when dissolved in water.
Now, let's calculate the total molarity of all the ions present:
For CuSO4:
- The molarity of copper ions (Cu2+) is 0.1 M (as given).
- The molarity of sulfate ions (SO4^2-) is also 0.1 M (since there is one sulfate ion produced for each copper ion).
For Al2(SO4)3:
- The molarity of aluminum ions (Al^3+) is 2 × 0.1 M = 0.2 M (as there are two aluminum ions produced for each Al2(SO4)3).
- The molarity of sulfate ions (SO4^2-) is 3 × 0.1 M = 0.3 M (since there are three sulfate ions produced for each Al2(SO4)3).
Finally, let's calculate the total molarity:
Total molarity of copper ions (Cu2+) = 0.1 M
Total molarity of aluminum ions (Al^3+) = 0.2 M
Total molarity of sulfate ions (SO4^2-) = 0.1 M (from CuSO4) + 0.3 M (from Al2(SO4)3) = 0.4 M
Therefore, the total molarity of all the ions present in the solution is: 0.1 M Cu2+, 0.2 M Al^3+, and 0.4 M SO4^2-.
I assume this is 0.1M CuSO4 and 0.1M Al2(SO4)3 IN THE SAME SOLUTION. If they are prepared solutions that have been mixed it is calculated a different way.
So it will be 0.1M in Cu^2+.
0.2 M x 2 = Al^3+
0.1M + 0.3M = 0.4M in SO4^2-