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a) \(n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,1\cdot0,12+0,1\cdot0,04=0,016\)
\(C_M=\dfrac{0,016}{0,2}=0,08M\)
\(\Rightarrow pH=-log\left(0,08\right)=1,1\)
b) \(n_{OH^-}=n_{KOH}+2n_{Ba\left(OH\right)_2}=0,012+2\cdot0,004=0,02\)
\(C_M=\dfrac{0,02}{0,2}=0,1\)
\(\Rightarrow pH=-log\left(\dfrac{10^{-14}}{0,1}\right)=13\)
Ví dụ 5 :
n KOH = 0,02.0,35 = 0,007(mol)
n HCl = 0,08.0,1 = 0,008(mol)
$KOH + HCl \to KCl + H_2O$
n HCl pư = n KOH = 0,007(mol)
=> n HCl dư = 0,008 - 0,007 = 0,001(mol)
V dd = 0,02 + 0,08 = 0,1(mol)
=> [H+ ] = CM HCl dư = 0,001/0,1 = 0,01M
=> pH = -log(0,01) = 2
CNaOH sau = 0,01*100/(100+100)=0.005M
CKOH sau= 0,02*100/(100+100)=0,01M
NaOH →Na+ + OH-
0,005---------->0,005 (M)
KOH →K+ + OH-
0,01--------->0,01(M)
=> [OH-]=0.01+0.005=0.015=> [H+]=10-14:0,015=6,67.10-13 (M)
=> pH= -log(6,67.10-13)= 12,18
\(n_{NaOH}=0.1\cdot0.1=0.01\left(mol\right)\)
\(n_{KOH}=0.1\cdot0.1=0.01\left(mol\right)\)
\(V=0.1+0.1=0.2\left(l\right)\)
\(\left[Na^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)
\(\left[K^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)
\(\left[OH^-\right]=\dfrac{0.01+0.01}{0.2}=0.1\left(M\right)\)
\(b.\)
\(pH=14+log\left[OH^-\right]=14+log\left(0.1\right)=13\)
\(c.\)
\(H^++OH^-\rightarrow H_2O\)
\(0.02........0.02\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.02}{1}=0.02\left(l\right)\)
\(a.\)
\(n_{NaOH}=0.1\cdot0.1=0.01\left(mol\right)\)
\(n_{KOH}=0.1\cdot0.1=0.01\left(mol\right)\)
\(V=0.1+0.1=0.2\left(l\right)\)
\(\left[Na^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)
\(\left[K^+\right]=\dfrac{0.01}{0.2}=0.05\left(M\right)\)
\(\left[OH^+\right]=\dfrac{0.01+0.01}{0.2}=0.1\left(M\right)\)
\(b.\)
\(pH=14+log\left(0.1\right)=13\)
\(c.\)
\(H^++OH^-\rightarrow H_2O\)
\(0.02.......0.02\)
\(V_{H_2SO_4}=\dfrac{0.02}{1}=0.02\left(l\right)\)
a) Ta có: \(n_{NaOH}=0,1\cdot0,1=n_{KOH}=0,01\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}n_{OH^-}=0,02\left(mol\right)\\n_{Na^+}=n_{K^+}=0,01\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[OH^-\right]=\dfrac{0,02}{0,2}=0,1\left(M\right)\\\left[Na^+\right]=\left[K^+\right]=\dfrac{0,01}{0,2}=0,05\left(M\right)\end{matrix}\right.\)
b) Ta có: \(pH=14+log\left[OH^-\right]=13\)
c) PT ion: \(OH^-+H^+\rightarrow H_2O\)
Theo PT ion: \(n_{H^+}=n_{OH^-}=0,02\left(mol\right)\)
\(\Rightarrow n_{H_2SO_4}=0,01\left(mol\right)\) \(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,01}{1}=0,01\left(l\right)=10\left(ml\right)\)
1. Số mol OH- = 10-4.0,1 + 10-2.0,1 = 0,00101 mol.
=> nồng độ OH- = 0,00101: 0,2 = 0,00505M
=> pOH = - lg(0,00505) = 2,3 => pH = 11,7.
2. Sai đề: pH của NaOH = 1??????
Đáp án C
nOH- = 0,03 mol; nH+ = 0,1.10-1 = 0,01 mol
H+ + OH- → H2O
0,01 0,03 mol
nOH- dư = 0,02 mol; [OH-]dư= 0,02/0,2 = 0,1M, [H+] = 10-13 M, pH = 13
\(pH=10\)
\(\Rightarrow\left[H^+\right]=10^{-10}\)
\(\Rightarrow\left[OH^-\right]=10^{-4}\)
\(n_{OH^-}=10^{-4}.0,1=10^{-5}\left(mol\right)\)
\(n_{H^+}=0,1.2.0,01=0,003\left(mol\right)\)
\(\Rightarrow n_{H^+dư}=2,99.10^{-3}\left(mol\right)\)
\(\Rightarrow\left[H^+_{dư}\right]=\dfrac{2,99.10^{-3}}{0,2}=0,01495M\)
\(\Rightarrow pH\approx1,83\)