30.0 cm^3 of 0.1M KOH is completely neutralized by 20.0cm^3 of dilute H2SO4 to form K2SO4 solution.What is the molarity of the salt solution obtained?
Please explain
Concentration of salts
2012-02-16 7:12 am
回答 (2)
2012-02-16 8:56 am
✔ 最佳答案
2KOH+ H2SO4 → K2SO4 + H2ONo. of moles of KOH used = 0.1 x (30/1000) = 0.003 mol
No. of moles of K2SO4 formed = 0.003 x (1/2) = 0.0015 mol
Volume of the final solution = (30 + 20) cm³ = 0.05 dm³
Molarity of K2SO4 = 0.0015/0.05 = 0.03 M
參考: micatkie
2012-02-16 7:59 am
2KOH +H2SO4 -----> K2SO4 +2H2O
20cm3 H2SO4 ---->40CM3 KOH
ONLY USE 30CM3 KOH =15CM H2SO4
K2SO4 : (30/1000)x0.1= 3x10^-3 mol
Volume=50 cm3 molarity of the salt solution: (3x10^-3) / (50/1000)
=0.06 moldm-3
20cm3 H2SO4 ---->40CM3 KOH
ONLY USE 30CM3 KOH =15CM H2SO4
K2SO4 : (30/1000)x0.1= 3x10^-3 mol
Volume=50 cm3 molarity of the salt solution: (3x10^-3) / (50/1000)
=0.06 moldm-3
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