Question

[Mai phương thuý]

Cho 200g dung dịch NaOH 2% (d=1g/ml) vào 50g dung dịch H2SO4 49%. (d=1,05g/ml). Tính nồng độ mol của các ion

Answer 1

[Minh Nhân]

\(n_{NaOH}=\dfrac{200\cdot2\%}{40}=0.1\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{50\cdot49\%}{98}=0.25\left(mol\right)\)

\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

\(0.1...............0.05...........0.05\)

\(n_{Na_2SO_4}=0.05\left(mol\right)\)

\(n_{H_2SO_4\left(dư\right)}=0.25-0.05=0.2\left(mol\right)\)

\(V_{dd}=\dfrac{200}{1}+\dfrac{50}{1.05}=247.6\left(ml\right)=0.2476\left(l\right)\)

\(\left[Na^+\right]=\dfrac{0.05\cdot2}{0.2476}=0.4\left(M\right)\)

\(\left[H^+\right]=\dfrac{0.2\cdot2}{0.2476}=1.6\left(M\right)\)

\(\left[SO_4^{2-}\right]=\dfrac{0.05+0.2}{0.2476}=1\left(M\right)\)

Answer 2

[hnamyuh]

$n_{NaOH} = \dfrac{200.2\%}{40} = 0,1(mol)$
$n_{H_2SO_4} = \dfrac{50.49\%}{98} = 0,25(mol)$
$2NaOH + H_2SO_4 \to Na_2SO_4 + 2H_2O$
$n_{NaOH} : 2 < n_{H_2SO_4} : 1$ nên $H_2SO_4$ dư

$n_{H_2SO_4\ dư} = 0,25 - 0,1.0,5 = 0,2(mol)$

$n_{H^+\ dư} = 0,2.2 = 0,4(mol)$

Sau phản ứng :

$V_{dd} = \dfrac{200}{1} + 50.1,05 = 252,5(ml) = 0,2525(lít)$

Bảo toàn Na, S ta có : 

$[Na^+] = \dfrac{0,1}{0,2525} = 0,4M$
$[SO_4^{2-}] = \dfrac{0,25}{0,2525} = 0,99M$
$[H^+] = \dfrac{0,4}{0,2525} = 1,58M$