A 0.100 M solution of K2SO4 would contain the same total ion concentration as which of the following solutions?
A. 0.0800 M Na2CO3
B. 0.100 M NaCl
C. 0.0750 M Na3PO4
D. 0.0500 M NaOH
.0750
C. 0.0750 M Na3PO4.
You can trust me.
Hmm, let me see if I can "solve" this question with a touch of humor.
Well, it seems like we're dealing with some chemistry here! Now, if I were to become a chemist, I'd be the laughing stock of the lab. They'd probably call me the Clown Chemist. But enough about my hypothetical career choices, let's get back to the question!
We're comparing the total ion concentration of a 0.100 M solution of K2SO4 with some other solutions. So, which solution has the same total ion concentration?
In other words, we need to find the solution that has the same amount of ions floating around. It's like trying to find a chemistry match made in heaven!
Now, let's take a closer look at each option.
Option A is a 0.0800 M solution of Na2CO3. Hmm, interesting, but are the ions getting along swimmingly?
Option B offers us a 0.100 M solution of NaCl. Ah, the classic salty choice. But is it a total ion party?
Option C presents a 0.0750 M solution of Na3PO4. Now, that's a mouthful! But do the ions in this solution fit the bill?
Finally, option D is a 0.0500 M solution of NaOH. Ah, an alkaline contender! But is it enough to match the K2SO4 gang?
Drumroll, please! The answer is...option B! Yes, the 0.100 M solution of NaCl is the one that matches the total ion concentration of the K2SO4 solution.
So, congrats to option B for finding its chemistry soulmate! I hope they live happily ever after, surrounded by ions and laughter.
To determine which solution has the same total ion concentration as a 0.100 M solution of K2SO4, we need to calculate the total number of ions in each solution.
Let's start by calculating the total number of ions in a 0.100 M K2SO4 solution. K2SO4 dissociates into three ions: two K+ ions and one SO4^2- ion.
Total number of ions = (concentration of K+ ions × number of K+ ions) + (concentration of SO4^2- ions × number of SO4^2- ions)
= (0.100 M × 2) + (0.100 M × 1)
= 0.200 + 0.100
= 0.300
Now, let's calculate the total number of ions in each given solution:
A. 0.0800 M Na2CO3: Na2CO3 dissociates into three ions: two Na+ ions and one CO3^2- ion.
Total number of ions = (0.0800 M × 2) + (0.0800 M × 1)
= 0.160 + 0.080
= 0.240
B. 0.100 M NaCl: NaCl dissociates into two ions: one Na+ ion and one Cl- ion.
Total number of ions = (0.100 M × 1) + (0.100 M × 1)
= 0.100 + 0.100
= 0.200
C. 0.0750 M Na3PO4: Na3PO4 dissociates into four ions: three Na+ ions and one PO4^3- ion.
Total number of ions = (0.0750 M × 3) + (0.0750 M × 1)
= 0.225 + 0.075
= 0.300
D. 0.0500 M NaOH: NaOH dissociates into two ions: one Na+ ion and one OH- ion.
Total number of ions = (0.0500 M × 1) + (0.0500 M × 1)
= 0.050 + 0.050
= 0.100
Based on the calculations, the solution with the same total ion concentration as the 0.100 M K2SO4 solution is solution C, which is a 0.0750 M Na3PO4 solution since both have a total ion concentration of 0.300. Therefore, the answer is C.
3ewe
K2SO4 split when it ionizes as follows:
K2SO4(aq) ==> 2K^+ + SO4^2- or a total of three ions for each mol K2SO4. The concn in the problem for K2SO4 is 0.1 so the total concn of ions = 0.3M
Go through the answers and see which is larger.
Na2SO4 = 3 ions x 0.08 = 2.4M ions
NaCl = 2 ions x 0.1 = 0.2 M ion.
etc.