What is the final [Na+] in a solution prepared by mixing 70.0 mL of 3.00 M Na2SO4 with 30.0 mL of 1.00 M NaCl?

### 1 Answer | Add Yours

Amount of 3M `Na_2SO_4` added `= 3/1000xx70`

Amount of 1M `NaCl` added `= 1/1000xx30`

When we mix the two solutions the new volume of the solution will be the total of two mix.

Volume of final solution = 70+30 = 100ml

`Na_2SO_4 rarr 2Na^++SO_4^(2-)`

`NaCl rarr Na^++Cl^-`

Mole ratio

`Na_2SO_4:Na^+ = 1:2`

`NaCl:Na^+ = 1:1`

Amount of `Na^+` in final mix `= 1/1000(2xx3xx70+30xx1)`

`[Na^+] = 1/1000(2xx3xx70+30xx1)xx1000/100 = 4.5 `

*SO the `[Na^+]` in the mix is 4.5M*

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes