a) Đặt \(n_{CH_3COOH}=x\left(mol\right)\)

\( n_{H^+}=10^{-2,88}\cdot0,05=6,59\cdot10^{-5}\)

\(CH_3COOH\text{ }⇌\text{ }CH_3COO^-\text{ }+\text{ }H^+\)

Bd \(x\)

PL \(6,59\cdot10^{-5}\)_\(6,59\cdot10^{-5}\)__\(6,59\cdot10^{-5}\)

Sau \(x-6,59\cdot10^{-5}\)_\(6,59\cdot10^{-5}\)_\(6,59\cdot10^{-5}\)

\(\Rightarrow K_a=\frac{6,59\cdot10^{-5}\cdot6,59\cdot10^{-5}}{x-6,59\cdot10^{-5}}=10^{-4,76}\\ \Rightarrow x=3,16\cdot10^{-4}\)

b) \(NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\)

\(6,32\cdot10^{-4}\)___\(6,32\cdot10^{-4}\)

\(\Rightarrow n_{NaOH\left(dư\right)}=4,368\cdot10^{-3}\\ \Rightarrow pH=12,46\)

Bài 1:

n_CH3COOH = 0,08.1,5 = 0,12 (mol)

PTHH: CH₃COOH + C₂H₅OH --H⁺,t^o--> CH₃COOC₂H₅ + H₂O

                0,12----------------------------->0,12

=> m_CH3COOC2H5 = 0,12.88 = 10,56 (g)

Bài 2:

n_CH3COOH = 2.0,1 = 0,2 (mol)

PTHH: 2CH₃COOH + Mg --> (CH₃COO)₂Mg + H₂

                  0,2------->0,1----------------------->0,1

=> m_Mg = 0,1.24 = 2,4 (g)

PTHH: C₂H₄ + H₂ --t^o,Ni--> C₂H₆

              0,1<--0,1

=> V_C2H4(đktc) = 0,1.22,4 = 2,24 (l)

Sau phản ứng, $V_{dd} = 1 + 1 = 2(lít)$
$n_{H^+\ dư} = 2.10^{-pH} = 0,2(mol)$
$n_{H^+\ ban\ đầu} = n_{HCl} = 1.2 = 2(mol)$
$\Rightarrow n_{NaOH} = n_{OH^-} = n_{H^+\ pư} = 2 - 0,2 = 1,8(mol)$
$\Rightarrow V_{dd\ NaOH} = \dfrac{1,8}{1,8} = 1(lít)$

Mk đang ko hiểu lắm phần vdd giải thích hộ mk với 

\(n_{CH_3COOH}=0.5\cdot1=0.5\left(mol\right)\)

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

\(0.5...........................................0.5\)

\(m_{CH_3COOH}=0.5\cdot60=30\left(g\right)\)

\(m_{CH_3COONa}=0.5\cdot82=41\left(g\right)\)

\(n_{Na_2CO_3}=0.2\cdot0.5=0.1\left(mol\right)\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

\(0.2.........................0.1\)

\(\Rightarrow CH_3COOHdư\)

\(n_{CO_2}=n_{Na_2CO_3}=0.1\left(mol\right)\)

\(V_{CO_2}=0.1\cdot22.4=2.24\left(l\right)\)

a)

$m_{dd} = 16 + 234 = 250(gam)$
$V_{dd} = \dfrac{250}{1,05} = 238(ml)$

$n_{NaOH} = \dfrac{16}{40} = 0,4(mol)$

Suy ra :

$C\%_{NaOH} = \dfrac{16}{250} = 6,4\%$

$C_{M_{NaOH}} = \dfrac{0,4}{0,238} = 1,68M$

b)

$C\%_{NaOH} = \dfrac{16+10}{250+10}.100\% = 10\%$