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a)
$NaOH + HCl \to NaCl + H_2O$
$OH^- + H^+ \to H_2O$
b)
$n_{HCl} = 0,1.0,01 = 0,001(mol)$
$n_{NaOH} = 0,2.0,5= 0,1(mol)$
$\Rightarrow$ NaOH dư, HCl hết
$n_{NaOH\ pư} = 0,001 \Rightarrow n_{NaOH\ dư} = 0,1 - 0,001 = 0,099(mol)$
$\Rightarrow [OH^-] = \dfrac{0,099}{0,1 + 0,2} = 0,33M$
$\Rightarrow pOH = -log(0,33) = 0,48 $
$pH = 14 - pOH = 14 - 0,48 = 13,52$
a) \(n_{NaOH}=0,2.1=0,2\left(mol\right)\); \(n_{HNO_3}=0,2.0,5=0,1\left(mol\right)\)
\(NaOH+HNO_3\rightarrow NaNO_3+H_2O\)
0,2.............0,1
Lập tỉ lệ : \(\dfrac{0,2}{1}>\dfrac{0,1}{1}\) => Sau phản ứng NaOH dư
Dung dịch D gồm NaNO3 và NaOH dư
\(n_{NaNO_3}=n_{HNO_3}=0,1\left(mol\right)\)
\(n_{NaOH\left(pứ\right)}=n_{HNO_3}=0,1\left(mol\right)\)
\(n_{NaOH\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)
Ion trong dung dịch D : Na+ , NO3-, OH-
\(\left[Na^+\right]=\dfrac{0,1+0,1}{0,2}=1M\)
\(\left[NO_3^-\right]=\dfrac{0,1}{0,2}=0,5M\)
\(\left[OH^-\right]=\dfrac{0,1}{0,2}=0,5M\)
b)Trong dung dịch D chỉ có NaOH dư phản ứng
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
0,1................0,05
=> \(V_{H_2SO_4}=\dfrac{0,05}{1}=0,05\left(l\right)\)
a, \(\left[Na^+\right]=0,1\)
\(\left[K^+\right]=0,1\)
\(\left[OH^-\right]=0,2\)
\(\left[SO_4^{2-}\right]=0,2\)
\(\left[H^+\right]=0,4\)
b, \(n_{H^+}=0,1.0,4=0,04\left(mol\right)\)
\(n_{OH^-}=0,1.0,2=0,02\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
\(\Rightarrow n_{H^+dư}=0,02\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{0,02}{200}=10^{-4}\)
\(\Rightarrow pH=4\)
$n_{NaOH} = n_{KOH} = 0,1.0,1 = 0,01(mol)$
$n_{H_2SO_4} = 0,02(mol)$
OH- + H+ → H2O
Bđ : 0,01...0,04..................(mol)
Pư : 0,01...0,01...................(mol)
Sau pư : 0......0,03...................(mol)
$V_{dd} = 0,1 + 0,1 = 0,2(lít)$
Vậy :
$[K^+] = [Na^+] = \dfrac{0,01}{0,2} = 0,05M$
$[H^+] = \dfrac{0,03}{0,2} = 0,15M$
$[SO_4^{2-}] = \dfrac{0,02}{0,2} = 0,1M$
b)
$pH = -log(0,15) = 0,824$
a, \(n_{NaOH}=0,15.0,2=0,03\left(mol\right)=n_{Na^+}=n_{OH^-}\)
\(n_{KOH}=0,15.0,2=0,03\left(mol\right)=n_{K^+}=n_{OH^-}\)
⇒ ΣnOH- = 0,03 + 0,03 = 0,06 (mol)
\(n_{HCl}=0,25.0,4=0,1\left(mol\right)=n_{H^+}=n_{Cl^-}\)
\(H^++OH^-\rightarrow H_2O\)
0,06____0,06 (mol) ⇒ nH+ dư = 0,1 - 0,06 = 0,04 (mol)
\(\left[Na^+\right]=\left[K^+\right]=\dfrac{0,03}{0,15+0,25}=0,075\left(M\right)\)
\(\left[H^+\right]=\dfrac{0,04}{0,15+0,25}=0,1\left(M\right)\)
\(\left[Cl^-\right]=\dfrac{0,1}{0,15+0,25}=0,25\left(M\right)\)
b, pH = -log[H+] = 1
a, \(n_{HCl}=0,1.0,2=0,02\left(mol\right)=n_{H^+}=n_{Cl^-}\)
\(n_{H_2SO_4}=0,1.0,2=0,02\left(mol\right)=n_{SO_4^{2-}}\) \(\Rightarrow n_{H^+}=2n_{H_2SO_4}=0,04\left(mol\right)\)
\(n_{NaOH}=0,3.0,4=0,12\left(mol\right)=n_{Na^+}=n_{OH^-}\)
\(\Rightarrow\sum n_{H^+}=0,02+0,04=0,06\left(mol\right)\)
\(H^++OH^-\rightarrow H_2O\)
0,06__0,06 (mol)
⇒ nOH- dư = 0,12 - 0,06 = 0,06 (mol)
\(\Rightarrow\left\{{}\begin{matrix}\left[Cl^-\right]=\dfrac{0,02}{0,1+0,3}=0,05\left(M\right)\\\left[SO_4^{2-}\right]=\dfrac{0,02}{0,1+0,3}=0,05\left(M\right)\\\left[Na^+\right]=\dfrac{0,12}{0,1+0,3}=0,3\left(M\right)\\\left[OH^-\right]=\dfrac{0,06}{0,1+0,3}=0,15\left(M\right)\end{matrix}\right.\)
b, pH = 14 - (-log[OH-]) ≃ 13,176
\(n_{OH^-}=n_{NaOH}=0,3.1,5=0,45\left(mol\right)\\ n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,2x+0,5.0,2.2=0,2x+0,2\left(mol\right)\)
PT ion rút gọn: \(H^++OH^-\rightarrow H_2O\)
0,45<---0,45
\(\Rightarrow0,2x+0,2=0,45\Leftrightarrow x=1,25M\)
Ta có: \(V_{dd}=0,3+0,2=0,5\left(l\right)\) và \(\left\{{}\begin{matrix}n_{Na^+}=0,45\left(mol\right)\\n_{Cl^-}=0,2.1,25=0,25\left(mol\right)\\n_{SO_4^{2-}}=0,2.0,5=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}C_{Na^+}=\dfrac{0,45}{0,5}=0,9M\\C_{Cl^-}=\dfrac{0,25}{0,5}=0,5M\\C_{SO_4^{2-}}=\dfrac{0,1}{0,5}=0,2M\end{matrix}\right.\)
a, \(NaHCO_3+NaOH\rightarrow Na_2CO_3+H_2O\)
\(HCO_3^-+OH^-\rightarrow CO_3^{2-}+H_2O\)
b, \(\left[Na^+\right]=\dfrac{0,2.1,5+0,12.1,6}{0,2+0,12}=1,5376M\)
\(\left[CO_3^{2-}\right]=\dfrac{0,2.1,5}{0,2+0,12}=0,9375M\)
\(n_{H^+}=0,3\left(mol\right)\)
\(n_{OH^-}=0,192\left(mol\right)\)
\(\Rightarrow n_{H^+dư}=0,108\left(mol\right)\)
\(\Rightarrow\left[H^+\right]=\dfrac{0,108}{0,2+0,12}=0,3375M\)
\(\Rightarrow pH\approx0,47\)