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nKOH = 0,0015 mol
nBa(OH)2 = 0,001 mol
=> n[OH-] = 0,0015+0,001.2=0,0035 mol
=> p[OH-] = 2,456
=> p[H+] = 14-p[OH-]=11,544
\(n_{H^+}=0.2\cdot0.01\cdot2=0.004\left(mol\right)\)
\(n_{OH^-}=0.1\cdot0.01=0.001\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
\(0.001.....0.001\)
\(n_{H^+\left(dư\right)}=0.004-0.001=0.003\left(mol\right)\)
\(pH=-log\left(H^+\right)=-log\left(\dfrac{0.003}{0.2+0.1}\right)=2\)
Có: \(n_{H^+}=2n_{H_2SO_4}=2.0,2.0,01=0,004\left(mol\right)\)
\(n_{OH^-}=0,1.0,01=0,001\left(mol\right)\)
PT ion: \(H^++OH^-\rightarrow H_2O\)
____0,004___0,001 (mol)
\(\Rightarrow n_{H^+\left(dư\right)}=0,003\left(mol\right)\)\\(\Rightarrow\left[H^+\right]=\dfrac{0,003}{0,3}=0,01\)
\(\Rightarrow pH=2\)
Bạn tham khảo nhé!
Ok, để thử coi chứ tui ngu hóa thấy mồ :(
a/ \(n_{NaOH}=0,2.0,1=0,02\left(mol\right)\)
\(NaOH\rightarrow Na^++OH^-\)
\(n_{Na^+}=n_{OH^-}=0,02\left(mol\right)\)
\(\Rightarrow C_{MNa^+}=\frac{0,02}{0,4+0,1}=0,04\left(mol/l\right)\)
\(n_{Ba\left(OH\right)_2}=0,3.0,4=0,12\left(mol\right)\)
\(Ba\left(OH\right)_2=Ba^{2+}+2OH^-\)
\(\Rightarrow n_{OH^-}=0,24\left(mol\right);n_{Ba^{2+}}=0,12\left(mol\right)\)
\(\Rightarrow C_{MBa^{2+}}=\frac{0,12}{0,5}=0,24\left(mol/l\right)\)
\(n_{OH^-}=0,02+0,24=0,26\left(mol\right)\)
\(\Rightarrow C_{MOH^-}=\frac{0,26}{0,5}=0,52\left(mol/l\right)\)
b/ \(n_{HCl}=0,2V\left(mol\right)\)
\(\Rightarrow n_{H^+}=n_{Cl^-}=0,2V\)
\(\Rightarrow C_{MCl^-}=\frac{0,2V}{2V}=0,1\left(mol/l\right)\)
\(n_{H_2SO_4}=0,3V\left(mol\right)=\frac{n_{H^+}}{2}=n_{SO_4^{2-}}\)
\(\Rightarrow C_{MSO_4^{2-}}=\frac{0,3V}{2V}=0,15\left(mol/l\right)\)
\(n_{H^+}=0,2V+0,6V=0,8V\left(mol\right)\)
\(\Rightarrow C_{MH^+}=\frac{0,8V}{2V}=0,4\left(mol/l\right)\)
Bác nào hảo tâm giúp em mấy câu còn lại chớ đến đây thì em chịu chết òi :(
\(n_{HCl}=0.1\cdot0.03=0.003\left(mol\right)\)
\(n_{NaOH}=0.1\cdot0.01=0.001\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
Lập tỉ lệ :
\(\dfrac{0.003}{1}>\dfrac{0.001}{1}\Rightarrow HCldư\)
\(n_{HCl\left(dư\right)}=0.003-0.001=0.002\left(mol\right)\)
\(\left[H^+\right]=\dfrac{0.002}{0.1+0.1}=0.01\)
\(pH=-log\left(0.01\right)=2\)
\(b.\)
\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
\(0.001..........0.002\)
\(V_{Ba\left(OH\right)_2}=\dfrac{0.001}{1}=0.001\left(l\right)\)
a, \(\left\{{}\begin{matrix}n_{Ba^{2+}}=4.10^{-3}\left(mol\right)\\n_{Na^+}=3.10^{-3}\left(mol\right)\\n_{OH^-}=0,011\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[Ba^{2+}\right]=\dfrac{4.10^{-3}}{0,2+0,3}=0,008M\\\left[Na^+\right]=\dfrac{3.10^{-3}}{0,2+0,3}=0,006M\\\left[OH^-\right]=\dfrac{0,011}{0,2+0,3}=0,022M\end{matrix}\right.\)
b, Để trung hòa dung dịch A thì:
\(n_{H^+}=n_{OH^-}\)
\(\Leftrightarrow0,01.V_{ddHCl}=\left(0,02.2+0,01\right).0,2\)
\(\Leftrightarrow V_{ddHCl}=1\left(l\right)\)
$n_{NaOH} = 0,001(mol)$
$n_{Ba(OH)_2} = 0,01.0,15 = 0,0015(mol)$
$NaOH \to Na^+ + OH^-$
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$
Ta có :
$n_{OH^-}= 0,001 + 0,0015.2 = 0,004(mol)$
$V_{dd} = 0,1 + 0,15 = 0,25(mol)$
$[OH^-] = \dfrac{0,004}{0,25} = 0,016M$
$pOH = -log(0,016) = 1,795 \Rightarrow pH = 14 - 1,795 = 12,205$
e cảm ơn ạ 🥰