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\(C_{M_{HCl}}=a\left(M\right),C_{M_{H_2SO_4}}=b\left(M\right)\)
\(n_{HCl}=a\left(mol\right),n_{H_2SO_4}=b\left(mol\right)\)
\(n_{NaOH}=0.4\cdot0.5=0.2\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
\(a..........a.........a\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)
\(2b............b..........b\)
\(n_{NaOH}=a+2b=0.2\left(mol\right)\left(1\right)\)
\(m_{muối}=58.5a+142b=12.95\left(g\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.1,b=0.05\)
\(\left[H^+\right]=0.1+0.05\cdot2=0.2\left(M\right)\)
\(\left[Cl^-\right]=0.1\left(M\right)\)
\(\left[SO_4^{2-}\right]=0.05\left(M\right)\)
\(b.\)
\(pH=-log\left(0.2\right)=0.7\)
Đặt x, y là số mol HCl và H 2 SO 4 trong 40 ml dung dịch A.
HCl + NaOH → NaCl + H 2 O
H 2 SO 4 + 2NaOH → Na 2 SO 4 + 2 H 2 O
Số mol NaOH: x + 2y = 1.60/1000 = 0,06 mol (1)
Khối lượng 2 muối : 58,5x + 142y = 3,76 (2)
Từ (1), (2), giải ra : x = 0,04 ; y = 0,01.
C M HCl = 0,04/0,04 = 1(mol/l)
C M H 2 SO 4 = 0,01/0,04 = 0,25 (mol/l)
Có \(m_M=m_{Na}+m_{Cl}+m_{SO_4}=1,38+m_{Cl}+m_{SO_4}=3,76\)
\(\Rightarrow m_{Cl}+m_{SO_4}=2,38\)
\(\Rightarrow35,5n_{HCl}+96n_{H_2SO_4}=2,38\)
Lại có : \(n_O=n_{NaOH}=0,06\left(mol\right)\)
BtH : \(n_H=2n_{H_2O}=2n_O=n_{H\left(NaOH\right)}+n_{H\left(HCl\right)}+2n_{H\left(H_2SO_4\right)}\)
\(\Rightarrow n_{HCl}+2n_{H_2SO_4}=0,06\)
\(\Rightarrow n_{HCl}=0,04\left(mol\right)\)
\(\Rightarrow C_{MHCl}=1M\)
\(H_2SO_4+2NaOH->Na_2SO_4+2H_2O\\ H_2SO_4+Na_2CO_3->Na_2SO_4+CO_2+H_2O\\ n_{Na_2CO_3}=0,1mol=n_{H_2SO_4dư}\\ n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}.0,15.2=0,15mol\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,15+0,1}{0,25}=1\left(M\right)\\ m_{Na_2SO_4}=142\left(0,15+0,1\right)=35,5g\)
PT: \(H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O\)
Ta có: \(n_{H_2SO_4}=0,25.1=0,25\left(mol\right)\)
a, Theo PT: \(n_{NaOH}=2n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow V_{NaOH}=\dfrac{0,5}{2}=0,25\left(l\right)\)
b, Theo PT: \(n_{Na_2SO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)
\(\Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,25}{0,25+0,25}=0,5\left(M\right)\)
$n_{NaOH} = 0,005(mol)$
$n_{Na_2CO_3\ sau\ pư} = n_{Na_2CO_3.10H_2O} = \dfrac{2,86}{286} = 0,01(mol)$
$NaHCO_3 + NaOH \to Na_2CO_3 + H_2O$
$n_{NaHCO_3} = n_{NaOH} = 0,005(mol)$
$n_{Na_2CO_3} = n_{Na_2CO_3\ sau\ pư} - n_{NaHCO_3} = 0,01 - 0,005 = 0,005(mol)$
$\Rightarrow C_{M_{NaHCO_3}} = C_{M_{Na_2CO_3}} = \dfrac{0,005}{0,02} = 0,25M$