Dung dịch A chứa H2SO4 0,5M ___________ B chứa NaOH 0,5M và Ba(OH)2 0,25M trộn 400ml dd A với 100ml dd B thu được 500ml dd C và m (g) kết tủa T a,Tính m b,tính nồng độ các chất trong C

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Loga Sinh Học lớp 12

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thuongthiet

Giải thích các bước giải:

a. 400 mL dung dịch A: $n_{H_2SO_4}=0,5.0,4=0,2\ mol$

100 mL dung dịch B: $n_{NaOH}=0,5.0,1=0,05\ mol; n_{Ba(OH)_2}=0,1.0,25=0,025\ mol$

$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$

$Ba(OH)_2+H_2SO_4\to BaSO_4+2H_2O$

Ta thấy: $n_{Ba^{2^+}}<n_{SO_4^{2^-}}$

$⇒n_{BaSO_4}=0,025\ mol⇒m=5,825\ mol$

b. Dd C chứa $Na_2SO_4, H_2SO_4$ dư

$n_{H_2SO_4\ du}=0,2-\dfrac{0,05}2-0,025=0,15\ mol; n_{Na_2SO_4}=\dfrac{0,05}2=0,025\ mol$

$⇒C_{M\ H_2SO_4}=\dfrac{0,15}{0,5}=0,3\ M;C_{M\ Na_2SO_4}=\dfrac{0,025}{0,5}=0,05\ M$

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hoangexpectgnar

Đáp án+Giải thích các bước giải:

`a)`

`n_{H_2SO_4} =0,4.0,5=0,2 mol`

`n_{Ba(OH)_2}=0,1.0,25=0,025 mol`

`n_{NaOH}=0,1.0,5=0,05 mol`

Thứ tự:

`H_2SO_4 + Ba(OH)_2 \to BaSO_4 + 2H_2O` `(1)`

`H_2SO_4 + 2NaOH \to Na_2SO_4 + 2H_2O` `(2)`

Theo pt:

`n_{H_2SO_4.pư}=n_{Ba(OH)_2 }+ 1/2 . n_{NaOH}`

`=0,025+1/2 . 0,05=0,05 mol<0,2`

`=> H_2SO_4` dư

`C: Na_2SO_4, H_2SO_4 `

Theo `(1):`

`n_{BaSO_4}=n_{Ba(OH)_2}=0,025 mol`

Khối lượng kết tủa:

`m_T=0,025.233=5,825g`

`b)`

`n_{H_2SO_4.dư}=0,2-0,05=0,15 mol`

Theo `(2):`

`n_{Na_2SO_4} =1/2 . n_{NaOH}=1/2 . 0,05=0,025 mol`

Nồng độ mol của `C` là:

`CM_{H_2SO_4}={0,15}/{0,5}=0,3M`

`CM_{Na_2SO_4}={0,025}/{0,5}=0,05M`

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