\(n_{NaOH}=0,1.0,01=0,001(mol)\\ \Rightarrow n_{OH^{-}}=0,001(mol)\\ n_{HCl}=0,03.0,2=0,006(mol)\\ \Rightarrow n_{H^{+}}=0,006(mol)\\ H^{+}+OH^{-}\to H_2O\\ 0,001<0,006\\ OH^{-} hêt; H^{+} dư\\ n_{H^{+}}=0,006-0,001=0,005(mol)\\ [H^{+}]=\frac{0,005}{0,1+0,2}=\frac{1}{60}M\\ \to pH=-log(\frac{1}{60})=1,77 \)

Bài 1:

Ta có: \(\Sigma n_{OH^-}=n_{NaOH}+2n_{Ba\left(OH\right)_2}=0,05.0,01+0,05.0,005.2=0,001\left(mol\right)\)

\(n_{H^+}=n_{HCl}=0,05.0,015=0,00075\left(mol\right)\)

PT ion: \(OH^-+H^+\rightarrow H_2O\)

______0,001__0,00075 (mol)

⇒ OH^- dư. n_{OH^- (dư)} = 2,5.10^-4 (mol)

\(\Rightarrow\left[OH^-\right]=\frac{2,5.10^{-4}}{0,1}=2,5.10^{-3}M\Rightarrow\left[H^+\right]=4.10^{-12}M\)

\(\Rightarrow pH\approx11,4\)

Bài 2: Đáp án D

Giải:

Ta có: \(\Sigma n_{H^+}=n_{HCl}+2n_{H_2SO_4}=0,1.0,002+0,2.2.x=2.10^{-4}+0,4x\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\frac{2.10^{-4}+0,4x}{0,3}M\)

\(\Rightarrow pH=-log\left(\frac{2.10^{-4}+0,4x}{0,3}\right)=2,7\)

\(\Rightarrow x\approx9,964.10^{-4}\approx10^{-3}\)

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 : 

\(\dfrac{0.003}{1}>\dfrac{0.001}{1}\Rightarrow HCldư\)

\(n_{HCl\left(dư\right)}=0.003-0.001=0.002\left(mol\right)\)

\(\left[H^+\right]=\dfrac{0.002}{0.1+0.1}=0.01\)

\(pH=-log\left(0.01\right)=2\)

\(b.\)

\(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)

\(0.001..........0.002\)

\(V_{Ba\left(OH\right)_2}=\dfrac{0.001}{1}=0.001\left(l\right)\)

$n_{NaOH} = 0,001(mol)$
$n_{Ba(OH)_2} = 0,01.0,15 = 0,0015(mol)$

$NaOH \to Na^+ + OH^-$
$Ba(OH)_2 \to Ba^{2+} + 2OH^-$

Ta có : 

$n_{OH^-}= 0,001 + 0,0015.2 = 0,004(mol)$
$V_{dd} = 0,1 + 0,15 = 0,25(mol)$
$[OH^-] = \dfrac{0,004}{0,25} = 0,016M$

$pOH = -log(0,016) = 1,795 \Rightarrow pH = 14 - 1,795 = 12,205$

\(n_{OH^-}=0,1.0,01+0,15.0,01.2=0,004\left(mol\right)\\ V_{ddsau}=100+150=250\left(ml\right)=0,25\left(l\right)\\ \Rightarrow\left[OH^-\right]=\dfrac{0,004}{0,25}=0,016\left(M\right)\\ pH=14+log\left[0,016\right]\approx12,204\)

\(n_{NaOH}=0,03.0,1=0,003\left(mol\right)\\ n_{HNO_3}=0,01.0,01=0,0001\left(mol\right)\\ NaOH+HNO_3\rightarrow NaNO_3+H_2O\\ Vì:\dfrac{0,0001}{1}< \dfrac{0,003}{1}\\ \Rightarrow NaOHdư\\ n_{NaOH\left(dư\right)}=0,003-0,0001=0,0029\left(mol\right)\\ \left[OH^-\left(dư\right)\right]=\left[NaOH_{dư}\right]=\dfrac{0,0029}{0,01+0,1}=\dfrac{29}{1100}\left(M\right)\\ \Rightarrow pH=14+log\left[\dfrac{29}{1100}\right]\approx12,421\)

a) ko bik ....

b) 13,5

c)2

d)2,4