mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2SO4

mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2SO4

 

 

Calculate the concentration of potassium ion (K^+) in grams per liter (L)

after mixing 100 mL of 0.250 M KCl and 200 mL of 0.100 M K2SO4.

 

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KCl(aq) → K^+(aq) + Cl^-(aq)

[K^+] = [KCl] = 0.250 M

 

(0.250 mol/L) (0.100 L) = 0.025 mol K^+

( 참고: MV=mol https://ywpop.tistory.com/7787 )

 

 

 

K2SO4(aq) → 2K^+(aq) + SO4^2-(aq)

[K^+] = 2[K2SO4] = 2(0.100 M) = 0.2 M

 

(0.2 mol/L) (0.200 L) = 0.04 mol K^+

 

 

 

0.025 + 0.04 = 0.065 mol K^+

 

 

 

K의 몰질량 = 39.10 g/mol 이므로,

0.065 mol × (39.10 g/mol) = 2.5415 g K^+

( 참고: n=W/M https://ywpop.tistory.com/7738 )

 

 

 

grams per liter 단위로 계산하면,

2.5415 g / (0.100+0.200 L)

= 8.47 g/L

 

 

 

답: 8.47 g/L