Question

Concentration of Na+ in a solution prepared by mixing 30.00mL of 0.12M NaCl with 70mL of 0.15M Na2SO4??

Answer 1

70.0mL = 0.07L 

3M = xmoles / 0.07 

x = 0.21moles of Na2So4 

There are two moles of Na+ for every 1 mole of Na2SO4 therefore this will result in 

0.21 * 2 = 0.42moles of Na 

30.0mL = 0.03L 

1M = xmoles / 0.03 

x = 0.03moles of NaCl 

There is 1 mole of Na+ for every 1 mole of NaCl. Therefore 

0.03mol * 1 = 0.03moles of Na+ 

when we add the solutions together we will add the moles of Na+ and add the volumes. 

0.03moles + 0.42moles = 0.45moles of Na+ 

0.07L + 0.03L = 0.1L 

[Na+] = (0.45mol) / 0.1L 

[Na+] = 4.5M 

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Answer 2

na+4.5m is the answer for 30.00ml I can't do the process sorry I hope you understand